JL  JL*/~\.I  \ 


IN  MEMORIAM 
FLOR1AN  CAJORI 


I/ 


CORRELATED  MATHEMATICS 

FOR 

SECONDARY  SCHOOLS 


Part  I       Algebra  First  Course 

12mo,  283  pages,  $1.10  postpaid 

Part  II      Plane  Geometry 

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CORRELATED    MATHEMATICS    FOR    SECONDARY    SCHOOLS 


PLANE  GEOMETRY 


BY 

EDITH   LONG 

DEPARTMENT   OF   MATHEMATICS,    HIGH   SCHOOL,    LINCOLN,    NEBRASKA 

AND 

W.  C.  BRENKE 

PROFESSOR   OF   MATHEMATICS   IN   THE   UNIVERSITY   OF   NEBRASKA 


NEW  YORK 

THE    CENTURY   CO, 

1916 


COPYRIGHT,  1916,  BY 
THE  CENTURY  CO. 


o  o 


2-4 


PREFACE 


IN  this  text,  as  in  the  "Algebra,"  the  authors  have  at- 
tempted to  present  the  subject-matter  in  a  way  more  suitable 
to  beginners  than  is  the  case  in  most  of  the  modern  books. 
Brief  and  concise  methods  of  exposition  are  therefore  largely 
avoided,  especially  at  the  start,  and  much  space  is  given  to 
the  explanation  and  analysis  of  theorems  and  problems,  in 
order  to  bring  out  clearly  the  plan  of  attack  and  the  method 
of  proof. 

In  the  presentation  of  theorems,  where  complete  proofs 
are  given,  the  following  plan  has  been  adopted: 

1.  Statement  of  theorem. 

2.  Figure. 

3.  Statement  of  what  is  given.     (Hypothesis.) 

4.  Statement  of  what  is  to  be  proved.     (Conclusion.) 

5.  Analysis. 

6.  Proof. 

Special  attention  is  called  to  the  analyses.  If  the  student 
can  be  taught  to  analyze  a  problem  clearly,  he  has  taken  a 
long  step  in  its  solution.  It  is  at  this  point  that  the  logical 
faculties,  and  the  ability  to  co-ordinate  and  apply  informa- 
tion previously  acquired,  receive  their  chief  development, 
and  too  much  emphasis  can  hardly  be  placed  on  this  point. 
The  proof  then  consists  merely  in  establishing  the  steps 
marked  out  in  the  analysis. 

In  a  number  of  theorems  and  problems,  only  the  state- 
ment and  analysis  are  given;  in  others  the  statement  alone. 
These  should  be  assigned  as  exercises,  the  student  being 
required  to  work  out  complete  proofs  and  to  file  them  in  a 
carefully  kept  note-book. 

There  is  no  better  way  to  vivify  the  treatment  of  Algebra 
than  through  geometric  interpretation  of  its  magnitudes; 
likewise,  Geometry  can,  and  does,  borrow  much  from  Algebra 


M306195 


vi  PREFACE 

in  using  its  rules  and  symbols  in  the  study  of  geometric 
forms.  The  authors  wish  to  emphasize  the  fact  that  in  this 
text  the  simpler  operations  of  Algebra  are  brought  often  into 
play,  so  that  the  student  is  made  to  utilize  the  facts  acquired 
in  his  first  years'  study  of  mathematics  to  aid  him  in  the 
work  of  the  second  year.  Important  illustrations  of  this 
will  be  found  in  the  geometric  treatment  of  factoring,  and 
in  the  presentation  of  the  subject  of  ratio  and  proportion, 
which  is  developed  both  algebraically  and  geometrically. 

The  lists  of  exercises  will  be  found  very  complete,  and  to 
contain  numerous  illustrations  of  the  practical  aspects  of 
Geometry.  No  attempt  should  be  made  to  cover  them  all 
in  one  year,  but  rather  to  select  and  choose  a  variety,  suffi- 
cient to  give  practice  and  to  illustrate  the  meaning  and  bear- 
ing of  the  text. 

There  is  perhaps  nothing  which  mars  the  usefulness  of 
the  study  of  Geometry  so  much  as  the  habit  of  making 
careless  and  slipshod  constructions.  The  teacher  is  urged 
to  insist  on  the  use  of  a  sharp  pencil,  not  too  soft,  and  of 
rule  and  compass,  for  the  latter  a  metal  point  attached  to 
the  pencil  if  nothing  better  is  available,  and  to  have  draw- 
ings made  to  exact  scale  and  measured,  whenever  possible. 
Much  work  of  this  sort  is  called  for  in  the  exercises. 

Some  of  the  results  of  careless  drawing  are  shown  in  the 
paradoxes  at  the  end  of  the  book.  It  will  be  well  to  present 
these  to  the  class  when  suitable  places  are  reached  in  the  text. 

The  discussion  of  incommensurable  cases  also  has  been 
placed  at  the  end  of  the  book,  after  some  study  of  the  notion 
of  limits.  They  may  be  taken  up  at  proper  points  in  the 
text,  or,  better,  at  the  end  of  the  course. 

References  to  the  authors'  text-book  on  Algebra  are  made 
under  the  designation  "First  Course."  Most  of  the  subject- 
matter  referred  to  may  be  found  in  any  other  modern  text. 

E.  LONG, 

LINCOLN,  NEBRASKA,  TTT      _,     ._ 

May,  1916  ,  W.    C.    BRENKE. 


CONTENTS 

LIST  OF  SYMBOLS  FACING  PAGE  1 

CHAPTER  I 1-10 

PART  I.      HISTORICAL  INTRODUCTION 
PART  II.    GEOMETRIC  FORMS.     POSTULATES  OF 
THE  STRAIGHT  LINE.     AXIOMS 

CHAPTER  II ,,....        11-83 

PART  I.      ANGLES   ^ 

PART  II..  TRIANGLES 

PART  III.  TRANSVERSALS  AND  PARALLELS 

CHAPTER  III .     84-111 

EQUALITY  OF  FIGURES 

CHAPTER  IV 112-146 

PART  I.      CIRCLES.  CENTRAL  ANGLES.  CHORDS 
PART  II.    INSCRIBED  ANGLES 

CHAPTER  V 147-167 

PART  I.      Loci 

PART  II.    COORDINATE  GEOMETRY 

CHAPTER  VI 168-220 

PART  I.      RATIO  AND  PROPORTION 

PART  II.    SIMILAR  FIGURES 

PART  III.  TRIGONOMETRIC  FUNCTIONS 

CHAPTER  VII 221-269 

PART  I.      MENSURATION 

PART  II.    DIVISION  OF  THE  PERIGON 

PART  III.  INCOMMENSURABLE  GASES 

APPENDIX.     SOME  GEOMETRICAL  PARADOXES    .      .   270-273 
INDEX 275 

PROTRACTOR  Inside  of  Back  Cover 


SYMBOLS 

The  symbols  -f ,  — ,  X ,  -*- ,  are  used  as  in  Algebra. 

=  equals,  is  equal  to. 

T^  is  not  equal  to. 

<  is  less  than. 

>  is  greater  than. 

~  similar,  is  similar  to. 

^  similar  and  equal,  i.  e.  congruent. 

||  parallel,  is  parallel  to;  |  |s  parallels. 

J_  perpendicular,  is  perpendicular  to;  J§  perpendiculars. 
Z,  A  angle,  angles, 
rt.   Z  right  angle. 
A,  A  triangle,  triangles. 
O,  UJ  parallelogram,  parallelograms. 
CD, GO  rectangle,  rectangles. 
O,  ©circle,  circles. 

^arc. 

/.therefore. 

=  approaches. 


PLANE  GEOMETRY 


CHAPTER  I 

PART  I— HISTORICAL  INTRODUCTION.     PART  II— 

GEOMETRIC  FORMS.    POSTULATES  OF  THE 

STRAIGHT  LINE.    AXIOMS. 

PART  I— HISTORICAL  INTRODUCTION 

Before  beginning  the  study  of  geometry  it  will  be  well  for 
the  student  to  make  a  brief  survey  of  the  history  of  the 
development  of  the  subject,  that  he  may  approach  it  with 
his  mind  open  to  grasp  its  twofold  nature,  the  practical  and 
the  logical.  Many  of  the  truths  stated  in  the  geometry  are 
of  the  greatest  value  to  the  constructive  work  of  the  world, 
while,  from  the  educational  standpoint,  the  proof  that  the 
statements  made  are  true  trains  the  mind  in  clear  logical 
reasoning. 

On  the  practical  side,  the  pyramids  and  temples  of  Egypt 
tell  us  that  many  of  the  truths  of  geometry  were  known 
and  used  by  builders  thousands  of  years  before  Christ, 
while  a  papyrus  written  by  Ahmes  (now  preserved  in  the 
British  museum)  records  the  mathematics  known  to  the 
Egyptians  as  far  back  as  2500  B.  C.,  but  gives  no  intimation 
that  the  people  made  any  attempt  to  demonstrate  the  gen- 
eral truth  of  geometrical  theorems. 

To  the  Egyptian  geometry  was  of  practical  value.  It  re- 
mained for  the  Greeks  to  give  to  the  world  its  pure  logical 
reasoning. 

Thales  of  Miletus  lived  from  640  to  546  B.  C.  Being  a 
merchant,  his  commercial  pursuits  took  him  into  Egypt, 

1 


2  PLANE  GEOMETRY 

where  the  great  buildings  attracted  his  attention.  They 
had  attracted  the  attention  of  others  before  him  but  not  in 
the  same  way.  These  were  the  questions  that  bothered 
Thales:  How  was  it  possible  for  the  builders  to  place  them 
on  a  direct  north  and  south  line?  How  did  they  make  one 
side  of  the  base  at  exact 
right  angles  to  the  other? 
How  did  they  build  into 
the  air  with  such  perfect 
slope  and  symmetry?  He 
went  to  Egypt  as  a  mer- 
chant; he  remained  as  a 
student.  He  studied  with  the  priests  of  the  country  and 
soon  excelled  them  in  knowledge.  He  surprised  them  when 
by  a  simple  device  he  measured  the  heights  of  the  pyramids. 

When  he  had  learned  all  that  he  could,  he  returned  to 
Greece  where  he  gathered  about  him  scholarly  men  whom 
he  interested  in  the  subject,  and  formed  a  sort  of  club  to 
pursue  its  study.  This  group  of  scholars  is  known  in  history 
as  the  Ionian  school.  They  did  not  confine  their  attention 
to  mathematics  but  also  studied  philosophy  and  astronomy. 
In  mathematics  they  were  the  first  to  study  lines,  angles, 
and  surfaces,  apart  from  the  solids  on  which  they  found  them. 
The  truths  they  proved  were,  naturally,  of  the  simplest  nature. 

Thales  is  known  as  one  of  the  "seven  wise  men  of  Greece." 
He  died  about  546  B.  C. 

Some  years  before  the  death  of  Thales  another  famous 
school  was  started,  known  as  the  Pythagorean  school.  The 
leader  of  this  school  was  Pythagoras,  born  in  Samos,  580 
B.  C.  The  history  of  his  early  life  is  not  clearly  known  but 
it  is  thought  that  he  visited  Thales  and  his  school  and  upon 
advice  given  there  went  to  Egypt  and  Babylon  to  study. 

He  then  returned  to  Greece  where  he  tried  to  establish 
a  school  at  Samos,  his  birthplace.  This  was  not  a  success, 
so  he  went  to  Croton,  Italy.  Here  he  met  with  great  success. 


HISTORICAL  INTRODUCTION  3 

The  school  accomplished  many  great  things,  the  greatest 
of  which  was  to  make  the  study  of  geometry  a  form  of  liberal 
education.  The  plan  of  this  school  differed  greatly  from  that 
of  the  Ionian  school.  It  was  very  secret.  No  one  was 
allowed  to  divulge  anything  that  went  on.  Especially  was 
it  offensive  if  any  one  boasted  of  his  own  achievements  to 
those  outside.  Pythagoras  took  all  credit  to  himself,  al- 
though it  is  now  known  that  there  were  many  brilliant  minds 
connected  with  the  work.  This  mystery  was  the  cause  of 
an  attack  made  upon  the  school,  resulting  in  the  destruction 
of  the  buildings.  Pythagoras  fled,  but  was  pursued  and 
killed.  The  school  itself  lasted  for  about  200  years,  although 
it  did  not  continue  as  a  secret  brotherhood. 

Again  we  find  an  overlapping  of  schools.  Before  the 
Pythagoreans  had  entirely  disappeared,  there  arose  in  Athens 
a  group  of  men  known  as  the  Sophists,  "wise  men"  who 
taught  the  people  rhetoric,  philosophy,  and  mathematics. 
This  was  more  like  our  present  day  school,  since  the  leaders 
took  pay  for  their  work.  The  Pythagoreans  scorned  to 
take  money  for  imparting  knowledge. 

Somewhat  later  during  the  time  that  Athens  held  high 
place  in  the  literary  world,  Plato  founded  his  school  known 
as  the  Platonic  school.  Plato  was  a  philosopher  and  not  a 
professed  mathematician,  but  he  took  hold  of  the  study  and 
brought  to  it  carefully  stated  definitions,  analysis,  and  logi- 
cal methods  of  reasoning.  He  stimulated  the  study  by 
placing  over  the  entrance  of  his  school,  "Let  no  one  who  is 
unacquainted  with  geometry  enter  here."  Athens  was 
conquered  by  Philip  of  Macedon  and  her  great  power 
broken  in  338  B.  C. 

Shortly  after  the  destruction  of  Athens  our  history  takes 
us  back  to  Egypt.  Alexander  the  Great  founded  Alexandria, 
which  was  shortly  to  become  the  seat  of  the  greatest  univer- 
sity of  the  time,  known  as  the  First  Alexandrian  School. 
One  of  its  first  and  most  noted  teachers  of  mathematics  was 


4  PLANE  GEOMETRY 

Euclid.  Let  the  student  fix  this  name  well  in  mind  for  it 
was  Euclid  who  wrote  our  first  geometry.  He  lived  about 
300  B.  C.  It  has  been  stated  that  no  ancient  writer  in  any 
branch  of  knowledge  has  held  such  a  commanding  position 
in  modern  education  as  has  Euclid  in  his  "Elements  of 
Geometry."  He  gathered  the  work  of  the  preceding  cen- 
turies, completed  proofs,  added  new  proofs,  and  arranged 
the  whole  in  logical  order.  The  modern  texts,  such  as  you 
are  about  to  study,  are  largely  selected  from  this  work  of 
Euclid.  The  original  work  consisted  of  thirteen  chapters 
usually  spoken  of  as  books. 

A  man  who  has  given  us  much  in  mathematics  was  a 
student  in  the  Alexandrian  school.  This  man  was  Archimedes 
(287-212  B.  C.).  You  will  meet  with  his  name  often  in 
physics,  since  he  excelled  in  the  discoveries  of  physical 
laws.  However,  your  attention  will  be  called  especially  to 
him  when  you  take  up  the  study  of  solid  geometry. 

This  short  account  is  but  the  setting  of  the  history  of 
geometry.  As  the  work  develops,  attention  will  be  called 
to  the  work  done  in  the  individual  schools  as  far  as  it  is 
known.  From  this  we  hope  that  you  may  see  how  geometry 
began  in  Egypt,  practical  but  without  logical  reasoning, 
was  carried  to  the  Ionian  Isles  and  nourished,  grew  rapidly 
in  the  Pythagorean  school  in  Italy,  received  system  and 
logic  from  Plato  in  Athens,  and  was  carried  back  to  Egypt 
to  be  brought  into  a  state  of  high  perfection  by  Euclid. 


I,  §1] 


GEOMETRIC  FORMS 


PART  II— GEOMETRIC  FORMS. 
THE  STRAIGHT  LINE. 


POSTULATES  OF 
AXIOMS. 


1.  As  you  go  about  out  of  doors  notice  the  forms  of  objects, 
such  as  the  roofs  of  houses,  spires,  doors  and  windows  of 
churches,  trees  and  shrubs.  Make  groups  of  the  names  of 
these  objects  as  in  your  opinion  they  conform  in  a  general 
way  to  one  of  the  following  forms. 


Cube 


Prism 


Pyramid 


Frustum 
of  Pyramid 


Cylinder 


Cone 


Frustum 
of  Cone 


Sphere 


2.  It  is  easy  to  make  models  of  some  of  these  forms.  From 
stiff  paper  or  cardboard  cut  out  at  least  one  of  the  following 
patterns,  which  should  be  enlarged  four  or  five  times;  fold 
along  the  heavy  lines  and  use  the  flaps  marked  by  dotted 
lines  to  fasten  the  figure  together.  The  drawing  of  the 
enlarged  pattern  will  require  some  assistance  from  the 
instructor.  If  desired,  the  flaps  may  be  left  off  and  the 
model  fastened  together  with  gummed  paper  strips. 


PLANE  GEOMETRY 


[I,  §3 


The  student  should  also  make  some  pattern  of  his  own 
design. 

3.  Suppose  that  you  had  made  your  model  of  tin  instead 
of  paper  so  that  you  could  fill  it  with  water,  salt,  mercury, 
what  change  would  take  place  as  you  filled  it  with  each  in 
turn?    If  it  were  air  tight  and  you  should  pump  out  all  the 
air,  how  would  your  model  differ  from  what  it  was  before? 

4.  When  your  model  is  filled  with  air,  is  not  the  inner  air 
as  completely  separated  from  the  outer  air  as  in  the  other 
cases,  the  water,  salt,  or  mercury  is  separated  from  the 
outer  air?     The  weight  of  your  model  is  changed  but  there 
is  one  thing  that  is  not  changed,  and  that  is  the  shape.     Now 
if  you  can  imagine  the  cardboard  out  of  which  your  model  is 
made  to  become  thinner  and  thinner  until  it  is  so  thin  that 
you  cannot  see  it  at  all,  still  keeping  in  mind  a  body  of 


I,  §5] 


GEOMETRIC  FORMS 


definite  shape,  separated  from  everything  else,  you  have  in 
mind  what  is  called  a  geometric  solid.  That  which  you 
think  of  as  separating  this  geometric  solid  from  everything 
about  it  is  called  a  geometric  surface. 

5.  However,  your  figure  as  it  is,  is  a  geometric  solid,  pro- 
vided you  agree  to  disregard  everything  about  it  but  shape 
and  size.     Thus  all  objects  that  you  see  are  solids  as  well 
as  shapes  that  you  have  in  mind  and  cannot  see.     The  card- 
board used  to  make  your  pattern  is  solid  and  is  separated 
from  the  air  about  it  by  surface  which  is  neither  air  nor 
cardboard. 

6.  Examine  your  model;  you  will  see  that  the  surface  is 
divided  into  different  portions.    That  whiqh  divides  one  part 
of  surface  from  another  is  called  a  line.    Are  there  lines  on 
your  model?     Point  out  the  lines.     Again  these  lines  are 
divided  into  different  portions  by  means  of  points.     Are 
there  points  on  your  model?     Are  these  lines  and  points,  air, 
cardboard,  or  neither? 

7.  We  shall  call  the  surfaces  of  the  solids  faces.    The 
lines  where  the  faces  intersect  are  called  edges,  and  the 
point  where  the  edges  intersect  are  called  vertices  (corners) . 

Count  the  number  of  faces,  the  number  of  edges  and  the 
number  of  vertices  of  each  of  your  models,  and  make  a  table 
of  the  numbers. 


Faces 

Vertices 

Edges 

f  +  v  =  e  +  2 

6 

8 

12 

6  +  8  =  12  +  2 

Do  you  find  that  the  same  relation  exists  between  the 


8  PLANE  GEOMETRY  U,§8 

numbers  in  each  case?  If  you  let  /  stand  for  the  number  of 
faces,  and  v  for  the  number  of  vertices,  and  e  for  the  number 
of  edges,  you  should  find  this  to  be  true : 

/  +  v  =  e  +  2. 

8.  It  was  stated  above  that  that  which  divides  one  portion 
of  surface  from  another  is  called  a  line.     Can  the  surface  of 
your  model  be  divided  by  other  lines  than  edges?     In  how 
many  different  ways  may  it  be  divided  by  different  lines? 
In  other  words,  how  many  lines  are  there  on  the  surface  of 
your  model?     Think  out  this  question  with  reference  to  the 
invisible  block  that  we  considered.    Can  these  lines  be  divided 
by  means  of  points  other  than  vertices?     How  many  points 
are  there  on  the  surface  of  your  model? 

9.  A  geometric  point  has  neither  length,   breadth,  nor 
thickness.    When  you  place  the  "point"  of  your  pencil  on 
a  paper  so  as  to  make  a  visible  mark,  have  you  made  a 
geometric  point?     Is  the  statement  true  that  between  two 
points  there  can  always  be  placed  an  unlimited  number  of 
other  points,  no  matter  how  close  together  the  points  are 
chosen? 

10.  Can  you  think  of  a  moving  point?     If  so,  moving  in 
how  many  different  directions?     How  far?     What  would 
you  suggest  as  the  name  of  the  path?     If  you  think  of  two 
fixed  points,  in  how  many  different  ways  could  a  third 
point  pass  from  one  to  the  other?     Draw  a  picture  of  some 
of  the  lines  generated  by  a  moving  point  passing  from  one 
fixed  point  to  another.    Of  these  lines  the  one  which  brings 
the  same  idea  to  our  minds  as  soon  as  its  name  is  mentioned 
is  called1  a  straight  line.     How  many  such  lines  can  be 
drawn  between  two  points?    What  are  some  of  the  properties 
of  a  straight  line?     Which  of  these  properties  do  the  two 
points  determine? 

In  how  many  points  can  two  straight  lines  intersect? 
The  above  discussion  brings  out  four  facts  whose  truth 


I,  §  10]  GEOMETRIC  FORMS  9 

we  assume,  and  which  will  be  needed  in  the  work  before  us. 
They  are  called  postulates  for  the  straight  line. 

Post.  I.  Between  two  points  only  one  straight  line  can  be 
drawn. 

Post.  II.     Two  points  determine  a  straight  line. 

Post.  III.  The  shortest  distance  between  two  points  is 
measured  on  the  straight  line  joining  them. 

Post.  IV.     Two  intersecting  lines  determine  a  point. 
In  geometry,  as  in  algebra,  we  shall  also  use  the  following 
assumptions,  called  axioms. 

Ax.  1.  Things  equal  to  the  same  thing  are  equal  to  each 
other. 

Ax.  2.     //  equals  are  added  to  equals,  the  sums  are  equal. 

Ax.  3.  //  equals  are  subtracted  from  equals,  the  remainders 
are  equal. 

Ax.  4.  //  equals  ore  multiplied  by  equals,  the  products  are 
equal. 

Ax.  5.  //  equals  are  divided  by  equals,  the  quotients  are 
equal. 

Ax.  6.  //  equals  are  added  to  unequals,  the  sums  are 
unequal  in  the  same  order. 

Ax.  7.  //  equals  are  subtracted  from  unequals,  the  re- 
mainders are  unequal  in  the  same  order. 

Ax.  8.  //  unequals  are  multiplied  by  positive  equals,  the 
products  are  unequal  in  the  same  order. 

Ax.  9.  //  unequals  are  divided  by  positive  equals,  the 
quotients  are  unequal  in  the  same  order. 

Ax.  10.  //  unequals  are  added  to  unequals,  greater  to 
greater  and  less  to  less,  the  sums  are  unequal  in  the  same  order. 

Ax.  11.     Like  powers  of  equals,  or  like  roots,  are  equal. 

Ax.  12.  //  three  magnitudes  are  such  that  the  first  is  greater 
than  the  second  and  the  second  greater  than  the  third,  then  the 
first  is  greater  than  the  third. 

Ax.  13.     The  whole  is  greater  than  any  of  its  parts. 

Ax.  14.     The  whole  is  equal  to  the  sum  of  all  its  parts. 


10 


PLANE  GEOMETRY 


[1, 1 11 


11.  Examining  various  geometric  solids  you  will  find  that 
some  have  flat  surfaces  and  others  have  curved  surfaces, 
while  some  have  both.  Those  which  have  curved  surfaces 
such  as  the  cylinder,  cone,  sphere,  are  frequently  referred 
to  as  the  round  bodies,  while  those  with  all  surfaces  flat  are 
called  polyhedra.  You  will  find  that  the  plane  surfaces  are 
of  various  shapes  of  which  the  following  are  the  most  common: 


Triangle  Quadrilateral  Pentagon  Hexagon 

Among  the  quadrilaterals  you  will  find  the  following: 


Rectangle 


Parallelogram 


Trapezoid 


In  your  homes  note  the  designs  of  wall-paper,  linoleum, 
rugs  and  tiling.  You  will  find  many  interesting  geometric 
shapes  among  them. 

REVIEW  OF  CHAPTER  I 

1.  Geometry  deals  primarily  with  form.     Name  the  common  forms. 
Do  things  that  grow  take  these  shapes?     What  parts  of  buildings  have 
these  various  forms? 

2.  Surfaces  are  either  curved  or  flat.     Name  solids  that  have  only 
flat  surfaces.     Name  solids  that  have  both  curved  and  flat  surfaces. 
Is  there  any  that  has  curved  surface  only? 

3.  What  forms  of  faces  are  found  on  the  solids  that  have  flat  sur- 
faces only?     What  forms  of  flat  faces  are  found  on  those  that  have 
both  flat  and  curved  surfaces? 

4.  The  objects  of  geometric  study  are  solids,  surfaces,  lines  and 
points. 

6.  State  the  postulates  for  the  straight  line. 

6.  State  the  axioms  in  algebraic  language,  as,  for  example,  Ax.  6: 
If  a  =  b  and  c  <  d,  then  a  +  c  <  b  +  d. 


CHAPTER  II 

PART  I— ANGLES.    PART  II— TRIANGLES.    PART  III— 
TRANSVERSALS  AND  PARALLELS.    SUMS  OF 
ANGLES  OF  POLYGONS.    PARALLEL- 
OGRAMS. 

PART  I— ANGLES 

12.  Angles — Definitions  and  Notation.  Suppose  that  we 
partially  open  a  fan,  or  two  of  the  arms  of  a  folding  ruler. 
Suppose  also  that  a  line  is  drawn  on  each  arm  of  the  ruler, 
starting  from  the  pivot  and  following  the 
middle  of  the  arm. 

Definitions.  The  figure  so  formed  by 
two  'straight  lines  starting  out  from  the 
same  point  is  called  an  angle.  The  point 
marked  A  in  the  figure  (the  pivot)  is  called 
the  vertex  of  the  angle.  The  lines  AB  and  AC  are  called 
the  arms  of  the  angle. 

Notation.  We  shall  often  use  a  single  letter,  usually  a 
capital,  placed  near  a  point  in  a  figure  to  designate  that 
point.  If  we  designate  a  certain  point  by  A,  and  another 
point  by  B,  the  straight  line  through  these  two  points  is 
called  the  line  AB,  or  the  line  BA.  We  would  say  "the  line 
A  B"  when  the  line  is  drawn  from  A  to  B;  we  would  say 
"the  line  B  A"  when  the  line  is  drawn  from  B  to  A.  Often 
it  makes  no  difference  which  we  use;  in  other  cases  a  dis- 
tinction is  necessary.  See  Ch.  V,  Part  II,  and  Ch.  VI, 
Prob.  I. 

The  distinction  is  of  great  importance  in  Algebra,  in  illus- 
trating the  idea  of  positive  and  negative  numbers. 

11 


12  PLANE  GEOMETRY  [II,  §  13 

To  designate  the  angle  formed  by  the  lines  AB  and  AC, 
we  say  "the  angle  BAG,"  or,  "the  angle  CAB."  The  first 
means  that  in  opening  up  the  angle  we  regard  AB  as  a  fixed 
arm  and  AC  as  revolving;  the  second  means  that  AC  is 
the  fixed  arm  and  that  AB  is  revolving.  Often  it  makes  no 
.difference  which  notation  is  used.  In  either  case,  to  desig- 
nate an  angle,  first  name  a  point  on  the  one  arm,  then 
name  the  vertex,  and  finally  name  a  point  on  the  other  arm. 

In  fixed  figures,  angles  may  usually  be  read  either  way. 

The  symbol  for  the  word  "angle"  is  Z ;  so  Z  B AC  means 
"angle  BAC." 

Exercise.  In  each  of  the  figures  below  read  off  the  points, 
lines,  and  angles  marked  in  it.  Draw  several  figures  of  your 
own,  letter  the  points  at  the  ends  of  each  line,  and  read 
the  angles. 


13.  Classification  of  Angles.  Angles  are  classified  accord- 
ing to  the  amount  of  turning  done  in  separating  the  arms 

In  the  figure  on  p.  13  suppose  AB  to  be  fixed  and  AC 
to  be  revolving;  when  AC  has  revolved  half  way  around,  so 
that  it  lies  just  opposite  to  AB  and  forms  one  straight  line 
with  it,  the  angle  .6 AC  is  called  a  straight  angle;  half  of  a 
straight  angle  is  a  right  angle;  if  AC  turns  through  less  than 
a  right  angle  it  forms  with  AB  an  acute  angle;  if  AC  turns 
through  more  than  a  right  angle  but  less  than  a  straight 
angle,  it  forms  with  A  B  an  obtuse  angle ;  more  than  a  straight 
angle  is  called  a  reflex  angle;  a  complete  turn  is  called  a 
perigon. 


II,  §  14] 


ANGLES 


13 


!\ 


C  A  B         A  B          A  B 

Straight  Angle  B  AC.         Right  Angle  B  AC  Acute  Angle  BAG. 


A  B 

Obtuse  Angle  BAG. 


Perigon  BAG. 


Reflex  Angle  BAG. 


Exercise.  Classify  each  of  the  angles  in  the  figures  on 
p.  12.  Draw  a  closed  figure  which  shall  contain  an  acute 
angle,  a  right  angle,  an  obtuse  angle,  and  a  reflex  angle. 

Definitions. 

A  polygon  is  a  figure  bounded  by  straight  lines. 

A  convex  polygon  is  one  whose  sides,  if  produced,  will  not 
cut  the  polygon. 

The  second  of  the  three  polygons  in  article  12  is  convex. 
The  third  is  concave. 

The  angle  HFM  is  called  a  re-entrant  angle. 

A  re-entrant  angle  is  always  greater  than  a  straight 
angle. 

14.  Measurement  of  Angles.  To  state  the  size  of  an 
angle  we  adopt  some  standard  angle  as  a  unit,  and  say  how 
many  of  these  units  are  needed  to  fill  the  given  angle.  Two 
different  units  are  in  common  use,  one  called  the  degree 
the  other  the  radian. 

Degree  Measure  of  Angles.  When  a  perigon  is  divided 
into  360  equal  parts,  each  such  part  is  called  a  degree. 


14 


PLANE  GEOMETRY 


[II,  §  14 


So  we  have 

360  degrees  =  a  perigon. 

Then  180  degrees  =  a  straight  angle 

and  90  degrees  =  a  right  angle. 

The  symbol  for  degrees  is  °,  so  that   10°  means   "ten 
degrees." 

Exercise.     How  many  degrees  in  each   of   the   following 
angles : 

(a)  One  fourth  of  a  right  angle.  (6)  Two  thirds  of  a  straight  an- 
gle, 
(c)  Two  fifths  of  a  perigon.         (d)  Seven  twelfths  of  a  perigon. 

Angles  are  usually  measured  with  a  protractor  (see  inside 
of  back  cover)  as  shown  in  the  figure  below.     To  measure 


Protractor 

a  reflex  angle,  measure  its  excess  over  a  straight  angle,  or 
measure  what  is  lacking  to  make  a  perigon. 

Exercise  1.  Draw  ten  different  angles,  some  acute,  some 
obtuse,  and  some  reflex.  Measure  each  and  write  its  value 
on  your  figure. 

Exercise  2.  Draw  a  triangle.  Measure  each  angle.  What  is 
their  sum?  Repeat  this  with  another  triangle  of  different  shape . 

Exercise  3.  Draw  a  triangle  and  tear 
apart  as  in  the  figure.  Place  the  three 
angles  with  their  vertices  together, 
one  angle  next  to  the  other,  without 
overlapping.  What  is  the  sum  of  the 
angles  of  the  triangle? 


II,  §  15]  ANGLES  15 

Exercise  4.  Repeat  Exercise  2,  using  a  quadrilateral, 
that  is,  a  figure  bounded  by  four  straight  lines.  Do  not 
draw  a  square  or  a  rectangle,  but  rather  a  figure  whose 
sides  and  angles  are  quite  unequal. 

Exercise  6.     Repeat  Exercise  3,  using  a  quadrilateral. 

Exercise  6.  Repeat  Exercises  2  and  3,  using  a  pentagon, 
that  is,  a  figure  bounded  by  five  straight  lines. 

Radian  Measure  of  Angles.  In  this  system  the  unit  of 
measure  is  a  radian,  instead  of  a  degree  as  in  the  system 
just  considered.  You  can  easily  make  a  protractor  gradu- 
ated in  radians. 

Exercise  1.  On  stiff  paper,  or,  better,  light  cardboard, 
draw  a  circle  with  a  radius  of,  say,  two  inches.  Carefully 
cut  it  out  and  mark  a  point  on  the  circumference  or  rim. 
On  a  good-sized  sheet  of  paper  draw  a  straight  line  and  mark 
off  on  it  parts,  each  equal  to  the  radius  of  the  circle.  Roll 
the  circle  carefully  along  this  line,  starting  with  the  marked 
point  on  the  rim  placed  at  the  beginning  of  the  first  division 
on  the  line.  Each  time  that  a  point  on  the  rim  of  the  rolling 
circle  reaches  a  division  point  on  the  line  mark  that  point 
on  the  rim.  Now  draw  lines  from  the  center  of  the  circle 
to  the  points  marked  on  the  rim.  You  then  have  a  series 
of  equal  angles,  each  of  which  is  one  radian. 

Exercise  2.     Define  a  radian. 

Exercise  3.  'By  rolling  the  circle  so  that  it  makes  just 
one  complete  turn,  find  approximately  how  many  radians 
there  are  in  a  perigon.  You  will  find  a  little  more  than 
six  radians.  Estimate  the  decimal  part  as  well  as  you 
can. 

15.  The  Number  TT.  The  number  of  radian  units  in  a 
perigon  is  not  a  whole  number,  as  you  found  in  the  last  exer- 
cise. Nor  can  this  number  be  expressed  either  by  a  fraction 
or  by  a  terminating  decimal.  It  is  a  so-called  incommen- 


16  PLANE  GEOMETRY  [II,  §  16 

surable  number  and  by  general  agreement  it  is  always  indi- 
cated by  2  TT,  TT  being  a  Greek  letter  called  "pi." 
We  therefore  have 

2  TT  radians  =  a  perigon  =  360  degrees. 
TT  radians  =  a  straight  angle  =  180  degrees. 

180 
Then  1  radian  =  -  -  degrees.     (About  57°.3). 

7T 

The  number  for  which  TT  stands,  to  four  decimal  places,  is 
3.1416;  less  exactly  it  is  ^f-.  It  should  be  remembered  that 
both  of  these  values  are  only  approximate. 

16.  Definitions.     When  an  angle  is  generated  by  the  turn- 
ing of  a  line  counter  clock- wise,  the  angle  is  said  to  be 
positive. 

When  an  angle  is  generated  by  the  turning  of  a  line  clock- 
wise, the  angle  is  said  to  be  negative. 

In  either  case,  the  first  position  of  the  moving  line  is  called 
the  initial  arm  of  the  angle,  and  the  last  position  is  called  the 
final  arm  of  the  angle. 

17.  Postulate  V.     All  straight  angles  are  equal. 
Corollary.     All  right  angles  are  equal. 

Note.    A  corollary  is  a  truth  that  follows  immediately  from 
what  precedes.      The   student  should   always  explain  the 
^connection. 

18.  Definition.     Two  angles  whose  algebraic  sum  is  equal 
to.  a  right  angle  are  called  complementary  angles. 

Two  complementary  angles  will  form  a  right  angle  if 
placed  with  their  vertices  together  and  with  the  initial  arm 
of  the  second  on  the  final  arm  of  the  first. 

When  both  angles  are  positive  and  are  placed  as  above, 
they  will  lie  adjacent  to  each  other  without  overlapping; 
each  of  them  is  less  than  90°,  and  hence  an  acute  angle. 
Of  two  complementary  angles  such  as  130°  and  —40°,  the 
second  will  lie  within  the  first. 


II,  §  19] 


ANGLES 


17 


Using  a  protractor  draw  the  complement  of  each  of  these 
angles : 


In  each  figure  suppose  the  final  arm  of  the  angle  to  be 
turned  until  it  makes  a  right  angle  with  the  initial  arm.  The 
angle  turned  through  is  the  complement  of  the  given  angle. 

19.  Definition.  Two  angles  whose  algebraic  sum  is  equal 
to  a  straight  angle  are  called  supplementary  angles. 

Two  supplementary  angles  will  form  a  straight  angle  if 
placed  with  their  vertices  together  and  with  the  initial  arm 
of  the  second  on  the  final  arm  of  the  first. 

Using  a  protractor  draw  the  supplement  of  each  of  these 
angles: 


20.  Definition.  Two  angles  whose  sum  is  a  perigon  are 
called  conjugate  angles. 

If  both  angles  are  positive,  as  130°  and  50°,  and  are 
placed  to  form  a  straight  angle,  they  will  lie  adjacent  to 
each  other  without  overlapping.  If  one  of  them  is  negative, 
as  230°  and  —  50°,  the  second  lies  within  the  first. 


18  PLANE  GEOMETRY  [II,  §  21 

Two  conjugate  angles  will  form  a  perigon  when  placed 
with  their  vertices  together  and  with  the  initial  arm  of  the 
second  on  the  final  arm  of  the  first. 

Using  a  protractor  draw  the  conjugate  of  each  of  the 
above  angles. 

21.  Exercises. 

1.  What  is  the  complement  of  an  angle  of  25°,  15°,  0°,  105°, 
115°,  173°,  -  36°,  -  128°,  a°,  i  TT  radians,  |  TT  radians,  -  J  TT 
radians. 

2.  What  is  the  supplement  of  each  of  the  angles  given  in 
Exercise  1? 

3.  What  is  the  conjugate  of  each  of  the  angles  given  in 
Exercise  1? 

4.  If  an  angle  is  2a  degrees,  what  is  its  complement? 

5.  If  an  angle  is  a  -f  3  degrees,  what  is  its  supplement? 

6.  If  an  angle  is  5 (a  —  2)  degrees,  what  is  its  conjugate? 

7.  If  an  angle  is  f  TT  +  5  radians,  what  is  its  complement? 

8.  If  an  angle  is  2a(—  3a  +  36)  degrees,  what  is  its  supple- 
ment? 

9.  If  an  angle  is  3  times  its  complement,  what  is  the  size 
of  the  angle? 

Solve  first  for  the  number  of  units  expressed  in  degrees, 
second,  for  the  number  expressed  in  radians.  (For  model  for 
solution,  see  First  Course,  pages  114,  115,  example  3.) 

10.  If  an  angle  is  2f  times  its  complement,  what  is  the  size 
of  the  angle? 

11.  If  an  angle  is  200°  more  than  2  times  its  complement, 
what  is  the  size  of  the  angle? 

12.  If  —  15°  be  added  to  }  of  the  supplement  of  an  angle, 
the  sum  will  be  equal  to  the  angle.    What  is  the  number  of 
degrees  in  the  angle? 


II,  §  22]  ANGLES  19 

13.  What  is  the  size  of  an  angle  if  its  complement  is  3  times 
its  supplement? 

14.  Two  angles  are  complementary.     If  10°  be  subtracted 
from  one  and  added  to  the  other,  the  two  angles  will  be  equal. 
What  is  the  number  of  degrees  in  each  angle? 

15.  Two  angles  are  conjugate.     If  J  TT  radians  be  added  to 
one  and  the  same  amount  subtracted  from  the  other,  the 
results  will  be  equal.    How  many  radians  are  there  in  each 
angle? 

22.  Theorem  I.    //  two  angles  are  equal,  their  complements 
are  equal. 


M 


We  have  given  the  equal  angles  ABC  and  MNK;  their 
complements  are  angles  STQ  and  HPR  respectively. 


r  s         P  H 

We  are  to  prove  that  angles  STQ  and  HPR  are  equal. 

Proof.  Subtract  Z  A  BC  from  a  right  angle,  and  subtract 
/.MNK  from  a  right  angle.  The  remainders  are  A  STQ 
and  HPR  respectively.  Why? 

But  these  remainders  are  equal  by  Ax.  3. 


20  PLANE  GEOMETRY  [II,  §  23 

The  following  algebraic  proof  of  this  simple  theorem  should 
be  studied  as  a  step  toward  more  difficult  proofs  of  later 
theorems. 

Let  us  say  that  there  are  a  degrees  in  the  Z.ABC;  then  there  are 
a  degrees  in  the  £MNK.  Why? 

Let  c  =  the  number  of  degrees  in  Z  STQ 

and  ci  =  the  number  of  degrees  in  Z  HPR. 

Then  a  +  c  =  90°.     Why? 

Also  a  +  ci=  90°.     Why? 

Therefore  a  +  c  =  a  +  d.     Why? 

Subtracting  a  from  both  sides  of  the  equation  we  have 

c  =  ci.     Why? 

Since  c  is  the  number  of  degrees  in  the  complement  of  angle  ABC  and 
d  is  the  number  of  degrees  in  the  complement  of  angle  MNK,  we 
have  proved  our  theorem.  State  it. 

23.  Theorem  II.     //  two  angles  are  equal,  their  supple- 
ments are  equal. 

Prove  this  as  in  the  preceding  theorem. 

24.  Theorem  III.     //  two  angles  are  equal,  their  conjugates 
are  equal. 

Prove  this  as  in  the  preceding  theorem. 

25.  Definition.    Vertical  Angles. 


Angle  ABE  and  angle  CBF  are  called  vertical  angles. 

Also  angle  FBA  and  angle  EEC  are  called  vertical  angles. 

Exercise.  Suppose  /.ABE  =  40°.  How  many  degrees  in 
Z.FBA1  Why?  Then  how  many  degrees  in  Z.CBF1  In 
Z  EBC  ?  Give  reason  for  each  answer.  Answer  the  same 
questions  when  /.ABE  =  n°.  What  do  you  conclude  about 
the  four  angles  at  B  ? 


II,  §  26]  ANGLES  21 

26.  Theorem  IV.     //  two  lines  intersect,  the  vertical  angles 
are  equal. 

We  have  given  in  the  figure  above  the  straight  lines  AC 
and  EF  intersecting  at  point  5,  forming  the  vertical  angles 
FBA  and  EEC. 

We  wish  to  prove  that  angle  FBA  equals  angle  EEC. 

Analysis.  What  is  the  sum  of  Z  FBA  and  Z  ABE,  no 
matter  what  the  number  of  angular  units  in  each  of  them? 
Give  reason.  Also  what  is  the  sum  of  Z  ABE  and  Z  EBCf 
Give  reason.  Can  you  form  an  equation  from  these  state- 
ments? Give  reason.  Can  you  subtract  the  same  angle 
from  each  member?  Why? 

Proof.  Does  this  prove  the  statement  to  be  proved? 
Write  the  answers  to  the  above  questions  in  good  English. 

Prove  the  other  two  vertical  angles  in  the  above  figure 
equal. 

27.  Exercises. 

1.  If  an  angle  is  bisected  and  a  line  is  drawn  through  the 
vertex  perpendicular  to  the  bisector,  it  makes  equal  angles 
with  the  arms  of  the  given  angle.     Prove  this  statement. 

2.  Find  the  value  of  the  angle  between  the  bisectors  of 
two  adjacent  complementary  angles. 

3.  Find  the  value  of  the  angle  between  the  bisectors  of  two 
adjacent  supplementary  angles. 

4.  Find  the  value  of  the  angle  between  the  bisectors  of 
two  adjacent  conjugate  angles. 

5.  How  do  the  bisectors  of  two  vertical  angles  stand  with 
relation  to  one  another?     Prove. 

6.  Suppose  /.FBA  in  the  figure  on  p.  20  to  be  divided 
into  five  equal  parts  by  lines  drawn  from  B.     If  these  lines 
are  produced  back  through  B,  show  that  /.EBC  will  be 
divided  into  five  equal  parts.     How  many  degrees  in  each  of 
these  parts  if  LCBV  =  60°? 


22  PLANE  GEOMETRY  m,  §  28 

7.  Draw  a  figure  like  that  on  page  20,  making  Z  CBF 
equal  to  50°.  Suppose  a  line  BX  drawn  in  the  angle  FBA, 
so  as  to  make  Z  FB X  equal  to  one  third  of  Z  FBA;  also 
a  line  B  Y  in  Z  C.BF,  so  as  to  make  /.  CBY  equal  to  one- 
third  of  Z  C£F.  Find  the  number  of  degrees  in  Z  YB  X. 
Check  result  by  measurement  of  your  drawing.  Solve  this 
exercise  when  Z  CBF  is  equal  to  n°. 


PART  II— TRIANGLES 

28.  Notation.  A  triangle  is  named  in  two  different  ways, 
either  by  placing  a  capital  letter  at  each  vertex  and  naming 
these  in  order,  preferably  counter-clockwise,  or  by  placing  a 
small  letter  on  each  side  and  naming,  preferably  counter- 
clockwise. Thus,  we  have  the  triangle  RMN  or  the  triangle 
rmn  (figure). 


If  both  vertices  and  sides  are  named  in  the  same  figure, 
the  letter  on  the  side  is  usually  the  small  letter  corresponding 
to  the  capital  letter  at  the  opposite  vertex.  It  does  not 
make  any  difference  in  naming  a  triangle  at  what  letter  you 
start. 

29.  Definition.  In  geometry  figures  which,  if  placed  to- 
gether, can  be  made  to  coincide,  are  called  congruent  figures. 

It  is  not  necessary  in  all  cases  that  the  figures  actually  be 
placed  one  on  the  other  in  order  that  we  may  know  that  they 
will  coincide  and  hence  are  congruent.  We  shall  now  try  to 
find  other  means  of  knowing. 


II,  §  30]  TRIANGLES  23 

30.  Theorem  V.  Two  triangles  are  congruent  if  they  have 
two  sides  and  the  included  angle  of  one  equal  respectively  to 
the  two  sides  and  the  included  angle  of  the  other. 


Given  triangles  ABC  and  AiBiCi,  such  that  side  a  =  side  ait 

side  6  =  side  61,  and  Z  C  =  Z  C\. 

i 
To  prove  triangle  ABC  congruent  to  triangle  A^BiCi. 

Analysis.  On  your  paper  draw  a  triangle  A  BC,  taking 
a  =  2  inches,  6=3  inches,  Z  C  =  40°.  Use  a  protractor  to 
lay  off  the  angle,  and  measure  off  a  and  b  on  the  arms  of  this 
angle. 

Draw  another  triangle  AiBiCi  with  the  same  values  for 
ai,  61,  and  Z  Ci  as  for  a,  b,  and  Z  C. 

We  wish  to  compare  these  triangles  as  to  form  and  size. 
To  do  this  cut  out  the  second  triangle  and  lay  it  on  the  first 
with  side  61  on  side  b  seeing  that  its  initial  point  C\  is  on  C, 
the  initial  point  of  b.  If  the  triangles  are  drawn  on  thin 
paper,  with  rather  heavy  lines,  you  can  lay  one  over  the 
other  without  cutting  out. 

Then  the  final  point  of  61  is  on  the  final  point  of  b.     Why? 

What  direction  will  the  line  ai  which  corresponds  to  a  take? 
Why? 

Where  will  the  final  point  of  ai  fall?     Why? 

Where  will  the  remaining  line  of  your  triangle  c\  fall? 
Why?  Art.  10,  Post.  I. 

Then  by  definition  these  two  triangles  are  congruent. 

The  symbolic  way  for  writing  this  fact  is 
A  ABC  &  A  AiBiCi. 


24  PLANE  GEOMETRY  pi,  §  31 

Proof.  Imagine  the  triangles  shown  in  the  book  to  be 
superposed  and  answer  the  questions  in  the  analysis.  There- 
fore the  triangles  coincide  throughout. 

A  ABC  &  A  A^d. 

State  the  theorem. 

Remark.  The  sign  of  congruency,  ii,  is  in  reality  two 
signs.  The  mark  above  is  an  s  written  sideways.  It  stands 
for  similar  and  means  that  the  figures  have  the  same  shape. 
The  equality  sign  below  the  s  means  that  the  figures  have  the 
same  size.  The  symbol,  then,  means  that  the  figures  have 
both  the  same  shape  and  the  same  size. 

31.  Theorem  VI.  Two  triangles  are  congruent  if  they  have 
two  angles  and  the  included  side  of  one,  equal  respectively  to 
the  two  angles  and  the  included  side  of  the  other. 


Given  A  ABC  and  A  AiBiCi  such  that  side  c  =  side  ci, 

Z  A  =  Z  Ai,  and  Z  B  =  Z  BL 

To  prove  triangle  ABC  congruent  to  triangle  AiBiCi. 

Analysis.  Draw  a  triangle  ABC,  taking  c  =  3  inches, 
Z  A  =  30°,  Z  B  =  50°.  To  do  this  first  draw  a  line  3 
inches  long  and  at  its  extremities  draw  Z  A  and  Z  B  using 
a  protractor.  Prolong  the  arms  of  these  angles  until  they 
meet. 

Draw  a  second  triangle  AiBiCi  with  side  Ci,  Z  AI,  and  Z  BI 
equal  to  side  c,  Z  A,  and  Z  B  of  the  first  triangle  respectively. 

Cut  out  the  second  triangle  and  place  it  on  the  first  with 
side  Ci  on  side  c,  point  AI  on  point  A.  Then  point  BI  falls 
on  point  B.  Why? 


II,  §  32]  TRIANGLES  25 

Side  fli  takes  the  direction  of  a.     Why? 

Side  61  takes  the  direction  of  6.     Why? 

Where  must  point  of  intersection  C\  fall.     Why? 

Proof.     Imagine  the  triangles  shown  in  the  book  to  be 
superposed  and  answer  the  questions  in  the  analysis. 

Therefore,  since  we  have  shown  that  the  two  triangles  co- 
incide throughout,  we  can  state  that 

A  AiBiCi  ^  A  ABC. 
State  the  proposition  proved. 

32.  Definitions. 

A  triangle  which  has  two  of  its  sides  equal  is  called  an 
isosceles  triangle. 

A  triangle  which  has  three  of  its  sides  equal  is  called  an 
equilateral  triangle. 

A  triangle  with  three  unequal  sides  is  called  scalene. 

33.  Theorem  VII.     The  angles  opposite  the  equal  sides  of 
an  isosceles  triangle  are  equal. 


We  are  given  the  triangle  ABC  with  the  side  a  equal  to 
side  b. 

We  are  to  prove  that  angle  BAG  equals  angle  CBA. 

Analysis.  The  only  relations  which  we  have  yet  proved 
about  triangles  are  the  truths  of  Theorems  V  and  VI.  (State 
them.)  So  it  will  be  necessary  to  have  two  triangles  that 
fulfil  one  of  these  relations.  These  triangles  must  be  such 
that  one  will  contain  one  of  the  angles  BAC  or  CBA,  and 
the  other  must  contain  the  other  of  these  angles.  Why? 


26 


PLANE  GEOMETRY 


[II,  §  34 


To  make  such  triangles  draw  a  line  bisecting  Z  ACB. 
(Use  protractor.)  Extend  this  bisector  until  it  intersects 
side  AB. 

You  thus  have  two  triangles  which  you  can  prove  con- 
gruent. What  has  this  to  do  with  proving  Z  BAC  equal  to 
Z  CBA1 


Proof.     In  A  ARC  and  RBC, 

side  b  =  side  a;  Why? 

side  CR  =  side  CR;  Why? 

Z  ACR  =  Z  £C£.  Why? 

What  can  you  say  about  A  ARC  and  RBCt  Why? 

What  can  you  say  about  A  RAC  and  GBR*!  Why? 

State  the  theorem  you  have  proved. 

Corollary.     The  angles  of  an  equilateral  triangle  are  equal. 
Why? 

Exercise.     Let  D  be  any  point  on  CR  in  the  figure  above. 
Draw  AD  and  BD.     Show  that  A  ABD  is  isosceles. 

34.  Definition.     One  line  is  said  to  be  perpendicular  to 
another  when  it  forms  a  right  angle  with  the  other. 

Corollary.     At  a  point  in  a  straight  line  only  one  perpen- 
dicular can  be  drawn  to  the  line. 


Exercise.     The  adjacent  figure 
shows  a  steel  square. 

Describe  a  steel  square. 
To  test  a  steel  square. 

Explain  how  a  steel  square  may 


II,  §  35]  TRIANGLES  27 

be  tested  by  the  method  suggested  in  the  following  figure. 
Line  AB  is  drawn  along  the  edge  of  the  tongue  with  the 
square  in  the  position  ABC,  with  the  blade  against  the 
line  MN.  Line  A\B  is  drawn  with  the  square  in  the  posi- 
tion AiBCi. 

What  is  wrong  with  this  square? 

A  A, 


Blade 


35.  Theorem  VIII.     Every  point  in  the  perpendicular  bi- 
sector of  a  straight  line  is  equidistant  from  the  ends  of  the  line. 

Suggestion.  Form  two  triangles  by  drawing  lines  from  the 
chosen  point  on  the  perpendicular  bisector  to  the  ends  of 
the  line.  Then  prove  the  triangles  congruent. 

36.  Suggestions.     The  student  should  observe  carefully 
the  steps  followed  in  the  proof  of  a  theorem.    They  are  as 
follows : 

(a)  Statement  of  the  theorem. 

(b)  Drawing  of  a  figure,  showing  the  things  spoken  of  in 
the  theorem. 

(c)  Statement  of  what  is  given,  called  the  hypothesis. 

(d)  Statement  of  what  is  to  be  proved,  called  the  con- 
clusion. 

(e)  Analysis,  indicating  the  method  of  proof. 

(f)  Proof. 

1.  Read  the  statement  of  the  theorem  carefully — two  or 
three  times  if  necessary  to  fix  it  clearly  in  your  mind. 

2.  Draw  a  figure  that  will  contain  each  thing  spoken  of  in 
the  statement  and  nothing  more. 


28 


PLANE  GEOMETRY 


[II,  §  36 


3.  Write  down  in  clear  statements  what  is  given  or  as- 
sumed to  be  true.     This  is  the  Hypothesis. 

4.  Write  down  what  you  are  to  prove.     This  is  the  Con- 
clusion. 

6.  Analyze,  thinking  carefully  on  the  following  questions: 

(a)  What  am  I  to  prove? 

(b)  In  order  to  prove  this  must  I  prove  two  things 
equal,  and  what  are  they? 

(c)  If  I  must  prove  two  things  equal,  can  I  find  two 
triangles  of  which  they  are  parts?     If  not,  can  I  make 
two  such  triangles  by  drawing  in  one  or  more  lines?    Can 
I  prove  these  triangles  congruent  according  to  Theorems 
V  or  VI? 

6.  Write  down  in  good  English  the  proof  of  the  theorem. 

In  proving  theorems,  be  careful  to  draw  a  general  figure, 
not  one  that  shows  merely  a  special  case.  For  example,  in 
a  general  theorem  about  a  triangle,  do  not  draw  a  right  trian- 
gle or  an  isosceles  triangle. 

Be  careful  not  to  jump  at  conclusions  from  a  mere  inspec- 
tion of  a  figure.  Thus,  by  inspection  of  the  figures  below, 
answer  the  following  questions: 

1.  Is  AB  parallel  to  CD? 

2.  Is  a  equal  to  6? 

Check  results  by  measurement. 


II,  §  37]  TRIANGLES  29 

37.  Exercises.     Geometric  and  Algebraic. 

1.  If  perpendiculars  are  drawn  at  the  ends  of  a  straight 
line  and  segments  are  cut  off  on  these  perpendiculars  by 
means  of  an  oblique  line  passing  through  the  mid-point  of 
the  first  line  drawn,  prove  that  the  segments  cut  on  the  per- 
pendiculars are  equal. 

2.  If  at  the  mid-point  of  the  side  c  of  the  triangle  ABC, 
of  which  a  and  b  are  equal  sides,  lines  are  drawn  making 
equal  angles  with  side  c  and  meeting  sides  a  and  6,  what  kind 
of  triangles  will  be  formed?     Prove. 

3.  If  the  equal  angles  of  an  isosceles  triangle  are  bisected 
and  the  bisectors  are  extended  until  they  intersect  the  op- 
posite sides  of  the  triangle  two  congruent  triangles  will  be 
formed.     Prove. 

4.  If  a  diagonal  bisects  the  opposite  angles  of  a  quadri- 
lateral, the  diagonal  divides  the  quadrilateral  into  two  con- 
gruent triangles.     Prove. 

5.  The  line  which  bisects  the  angle  between  the  equal  sides 
of  an  isosceles  triangle,  bisects  the  opposite  side  and  is  per- 
pendicular to  the  opposite  side.     Prove. 

6.  Draw  any  line  AB.     Draw  lines  AC  and  AD,  one  on 
one  side  of  AB  and  the  other  on  the  other  side,  making  equal 
angles  with  AB.     If  AC  and   AD  are  the  same  length, 
prove  that  BD  and  BC  are  the  same  length. 

7.  Draw  an  equilateral  triangle  ABC.     On  the  three  sides 
mark  P,  Q,  and  R,  respectively,  so  that  AP  equals  BQ  equals 
CR.     Prove  that  triangle  PQR  is  equilateral. 

8.  Draw  an  isosceles  triangle  ABC,  with  A  and  B  as  the 
vertices  of  the  equal  angles.     From  A  and  B  draw  perpendic- 
ulars to  the  opposite  sides.     Prove  that  these  perpendiculars 
are  equal. 

9.  Draw  an  equilateral  triangle  ABC;  extend  side  ABtoP, 
side  BC  to  Q  and  side  CA  to  R,  making  BP,  CQ,  and  AR 
all    equal    to    each   other.      Prove   that   triangle    PQR   is 
equilateral. 


30  PLANE  GEOMETRY  pi,  §  37 

10.  If  we  wish  to  measure  the  distance  from  A  to  B  and 
there  is  an  obstacle  in  the  way  to  prevent  the  direct  measure- 
ment, show  that  by  taking  the  following  measurements,  we 
can  determine  AB. 


Measure  AR,  R  being  any  point  that  can  be  reached  from 
both  A  and  B. 

Measure  RQ  equal  to  AR. 

Measure  BR}  and  then  measure  RP  equal  to  BR. 

Join  PQ  and  measure  it,  thus  finding  the  length  of  A  B. 
Explain. 

11.  Draw  an  isosceles  triangle  ABC  with  side  a  equal  to 
side  b.     Extend  side  c  each  way  through  the  vertices  to  points 
P  and  Q  respectively,  making  AP  equal  to  BQ.     Prove  that 
A  CPQ  is  isosceles. 

12.  An  isosceles  triangle  is  drawn.    Call  it  A  PQR,  p  and 
q  being  the  equal  sides.    Extend  side  q  through  the  point  R 
a  certain  length,  and  mark  the  point  where  you  stop  M. 
Extend  side  p  through  the  point  R  the  same  distance  that 
you  extended  q  and  mark  the  point  of  stopping  N.     Join  P 
to  M  and  Q  to  N  by  straight  lines.     Name  two  congruent 
triangles  and  prove. 

13.  Of  two  of  the  sides  of  a  triangle  one  is  1  unit  less  than 
2  times  the  other.    If  3  units  be  subtracted  from  the  one  and 
added  to  the  other  the  triangle  will  be  isosceles.     What  is 
the  length  of  each  of  the  two  sides? 


II,  §38]  TRIANGLES  31 

14.  Two  triangles  have  two  sides  of  one  equal  respectively 
to  two  sides  of  the  other,  but  the  included  angle  of  one  12 
degrees  more  than  twice  the  corresponding  angle  of  the  other. 
If  25  degrees  be  subtracted  from  the  larger  angle  and  47 
degrees  be  added  to  the  smaller  angle,  the  triangles  will  be 
congruent.     What  is  the  included  angle  of  each  triangle? 

15.  In  an  isosceles  triangle,  each  of  the  equal  angles  is 
twice  the  third  angle.     The  sum  of  the  three  angles  is  180°. 
What  are  the  angles? 

38.  Theorem  IX.  If  two  triangles  have  three  sides  of  the  one 
equal  respectively  to  three  sides  of  the  other,  the  triangles  are 
.  congruent. 


A, 


Given  two  triangles,  ABC  and  AiBiCi  having  side  ai  = 
side  a,  side  61  =  side  6,  side  c\  =  side  c. 

To  prove  triangle  ABC  congruent  to  triangle  AiBiCi. 

Analysis.  If  we  try  to  place  one  of  these  triangles  on  the 
other,  in  order  to  see  if  they  coincide,  as  we  did  in  theorems 
V  and  VI,  we  shall  not  be  able  to  determine  whether  the 
sides  coincide  or  not.  Try  placing  them  together  with  side  «i 
upon  a,  seeing  that  the  initial  points  fall  together.  Then 
the  final  points  will  fall  together.  Why?  Now  have  we  any 
way  of  knowing  whether  the  side  61  will  take  the  direction 
of  side  6?  Do  we  know  the  comparative  values  of  the 
angles  between  side  b  and  side  a,  and  between  side  61  and  side 
ai?  Therefore  can  we  tell  what  direction  61  will  take?  For 
this  reason  we  use  another  plan  of  placing  the  triangles 
together. 


32 


PLANE  GEOMETRY 


[II,  §  39 


Place  the  triangles  with  their  longest  sides  together,  that 
is  side  c\  upon  the  side  c,  but  have  the  two  triangles  on  op- 
posite sides  of  c. 


Why? 
Why? 

Why? 
Why? 
Why? 


Draw  the  line  CCi.     We  now  have  two  isosceles  triangles. 
Name  them.     From  these  we  can  show  an  equality  of  the 
angles  of  A  ABC  and  A  AiBid,  and  so  prove  the  two  tri- 
angles congruent  by  means  of  Theorem  V. 
Proof.  Z  dCB  =  Z  BdC. 

Z  ACCi  =  Z  CdA. 
therefore  Z  BCiA  =  Z  ACB. 

side  a  =  side  «i. 
side  b  =  side  61. 
What  parts  of  the  two  triangles  do  you  now  know  to  be 
equal?     Are  the  triangles  congruent?     Why? 
State  the  theorem  that  you  have  proved. 

39.  Exercises.     To  prove  the  following  exercises  proceed 
as  in  article  36,  adding  Theorem  IX  to  instruction  5,  (c) . 

1.  State  the  three  cases  of  congruency  of  triangles. 

2.  If    two    quadrilaterals 
have  four  sides  of  one  equal 
respectively  to  four  sides  of  the 
other  are  they  necessarily  con- 
gruent?    Take  four  strips  of 
paper  and  fasten  as  in  the  fig- 
ure.   Is  the  figure  rigid?    Do 


II,  §  39]  TRIANGLES  33 

the  same  with  three  strips.  Is  the  figure  rigid?  How  do  you 
account  for  the  difference?  What  strip  can  you  add  to  your 
quadrilateral  to  make  it  rigid? 

Now  state  a  theorem  for  congruency  of  quadrilaterals  and 
prove. 

3.  Note  the  adjacent  picture  of 
a  bridge  truss.     What  is  the  pur- 
pose of  the  diagonal  beams?  panei.tn.ss  Bridge. 

4.  Make  a  five-sided  figure  and  proceed  as  in  Exercise  2. 

5.  If  a  line  is  drawn  from  the  point  of  intersection  of  the 
equal  sides  of  an  isosceles  triangle  to  the  mid-point  of  the 
opposite  side,  it  is  perpendicular  to  the  opposite  side. 

6.  In  a  quadrilateral  A  BCD,  the  diagonals  AC  and  BD 
bisect  each  other.     Show  that  the   opposite   sides  of  the 
quadrilateral  are  equal.     Apply  theorems  IV  and  VI. 

7.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  either 
diagonal  divides  it  into  two  congruent  triangles. 

8.  If  two  equal  isosceles  triangles  are  placed  as  in  the 
figure  on  page  32,  show  that  line  CCi  divides  the  figure  into 
two  congruent  triangles. 

9.  Construct  an  isosceles  triangle  such  that  its  perimeter 
shall  be  8  inches  and  its  base  3  inches.     If  the  perimeter  of 
an  isosceles  triangle  is  n  inches,  how  large  can  the  base  be 
made  so  that  the  triangle  still  may  be  constructed. 

10.  Construct  a  triangle  such  that  its  perimeter  shall  be 
9  inches,  and  such  that  the  second  side  shall  be  twice  tLe 
first  side  and  that  the  third  side  shall  be  equal  to  the  sum  of 
the  other  two  sides. 

11.  Two  triangles  have  two  sides  of  the  one  equal  respect- 
ively to  two  sides  of  the  other,  but  the  third  side  of  the  first 
triangle  is  8  units  less  than  twice  third  side  of  the  second. 
If  the  two  triangles  are  to  be  made  congruent  without 
changing  the  lengths  of  the  equal  sides,  6  units  must  be 
taken  from  the  third  side  of  the  first.     Find  the  lengths  of 
the  third  side  of  each  at  present. 


34 


PLANE  GEOMETRY 


[II,  §  40 


12.  Pick  out  congruent  triangles  in  each  of  the  following 
designs.  State  why  congruent. 

In  the  first  design,  each  six-sided  figure  is  a  regular  hexa- 
gon, that  is,  a  figure  with  six  equal  sides  and  six  equal  angles. 


ill  z  I  ill 


TIIIIITI 


Tile  Field 


Tile  Border 


40.  Constructions.     We  shall  now  examine  methods  for 
constructing  geometric  figures  without  the  use  of  protractor 
and  measuring  line. 

For  this  purpose  you  will  need  a  compass  and  straight 
edge,  the  latter  preferably  not  marked  in  units. 

41.  Definitions. 

A  circumference  is  a  closed  curved  line  every  point  of 
which  is  at  the  same  distance  from  a  point  within  called  the 
center. 

A  part  of  a  circumference  is  called  an  arc. 

The  area  enclosed  by  a  circumference  is  called  a  circle. 

Any  straight  line  from  the  center  to  a  point  of  the  cir- 
cumference is  called  a  radius. 

42.  Assumptions.     We  shall  take  for  granted  the  following : 

1.  Two  circumferences  will  intersect  if  the  distance  between 
their  centers  is  less  than  the  sum  of  their  radii,  and  greater  than 
the  difference  of  their  radii. 

2.  The  radii  of  the  same  circle  or  of  equal  circles  are  equal. 

3.  If  a  straight  line  cuts  a  circumference  once,  it  cuts  it  twice. 


II,  §  43]  TRIANGLES  35 

4>  If  two  circumferences  cut  each  other  once,  they  cut  twice. 

5.  A  circumference  is  determined  when  its  center  and  radius 
are  known. 

43.  Problem  I.    Given  the  three  sides  of  a  triangle,  to  con- 
struct the  triangle. 


Given  the  three  line-segments  a,  b,  c. 
To  construct  the  triangle  abc. 

Analysis.  Suppose  the  work  done  and  that  A  ABC  has 
the  lines  a,  b,  c  for  its  sides.  We  see  that  if  A  B  is  the  side 
of  the  length  c,  B  will  be  one  of  the  vertices  and  A  another, 
while  the  vertex  C  lies  opposite  the  side  c.  Also  the  distance 
from  C  to  A  must  equal  b,  and  the  distance  from  B  to  C  must 
equal  a.  Now  every  point  at  distance  b  from  point  C  is  on 
the  circumference  drawn  about  C  as  center  and  with  a  radius 
b.  Every  point  at  distance  a  from  point  B  is  on  the  cir- 
cumference drawn  about  point  B  as  a  center,  with  radius  a. 
The  point  which  fulfils  both  of  these  conditions  is  on  both  of 
the  circumferences,  so  the  point  of  intersection  is  the  vertex  A. 

Construction.  Draw  a  straight  line.  On  this  lay  off  a 
distance  equal  to  segment  c.  Do  this  by  opening  the  com- 
pass points  so  that  they  will  just  span  segment  c,  and  then 
transfer  this  distance  to  the  line  you  have  drawn.  Mark 
one  end  B  and  the  other  A.  Now  set  your  compass  for 
length  b.  With  A  as  a  center  and  b  as  a  radius  draw  a  cir- 
cumference. In  like  manner  with  B  as  center  and  a  as 
radius,  draw  a  circumference. 


36  PLANE  GEOMETRY  HI,  §  44 

Call  the  points  of  intersection  C  and  Ci.  Join  these 
points  to  B  and  A,  and  you  have  two  triangles  which  fulfil 
the  required  conditions. 


Explain  why  all  other  triangles  which  have  these  sides 
are  the  same  size  and  shape  as  A  ABC. 

From  Assumption  1  of  article  42  and  the  above  discussion 
we  have  the  following  theorem: 

44.  Theorem  X.     The  sum  of  two  sides  of  a  triangle  must 
be  greater  than  the  third  side,  and  the  difference  must  be  less 
than  the  third  side. 

Draw  figure  and  explain. 

45.  Exercises.     Tell  whether  the  following  sets  of  numbers 
may  be  the  sides  of  triangles: 

1.  2,  4,  6.  3.  256,  174,  89. 

2.  5,  10,  3.  4.  .06,  .1,  .078. 
Problem  I  may  also  be  stated: 

46.  Problem  II.     To  construct  two  congruent  triangles  by 
making  their  corresponding  sides  equal. 

Let  the  student  supply  figure,  analysis,  construction,  and 
proof. 

Exercise.  The  sides  of  a  triangle  are  1.5  inches,  2  inches, 
and  3  inches  respectively.  Construct  a  triangle  congruent 
to  it.  Do  the  same  when  the  sides  of  the  given  triangle  are 
3,  4,  5  inches  respectively. 


II,  §  47]  TRIANGLES 

47.  Problem  III.     To  bisect  a  given  angle. 


37 


Given  angle  ABC. 
To  bisect  angle  ABC. 

Analysis.  Suppose  the  work  done  and  that  BD  is  the 
bisector,  D  being  any  point  on  it.  Suppose  that  we  take 
the  points  A  and  C  so  that  BA  equals  BC,  and  connect 
A  with  D,  and  C  with  D,  forming  two  triangles.  These 
triangles  are  congruent.  Prove  this.  Then  AD  and  DC 
are  equal.  Why?  Then  our  problem  resolves  itself  into-con- 
structing  two  triangles  with  the  three  sides  of  the  one  equal 
respectively  to  the  three  sides  of  the  other.  This  is  done  as 
in  Problem  II. 

Construction.  With  B  as  a  center  and  with  any  radius, 
lay  off  equal  distances  BA  and  BC. 

With  A  as  a  center  and  any  radius 
more  than  one-half  the  distance  AC, 
construct  an  arc. 

With  C  as  a  center  and  with  the  same 
radius,  construct  an  arc.  Call  the  point 
of  intersection  of  the  two  arcs  D. 

BD  is  the  bisector  of  Z  A  BC. 

Proof.     Give  the  complete  proof. 

Corollary  1.  At  a  given  point  on  a  straight  line  to  construct 
a  perpendicular  to  that  line. 

Construction.  Bisect  the  straight  angle  of  which  the  given 
point  is  the  vertex. 


38 


PLANE  GEOMETRY 


[II,  §  48 


Corollary  2.     From  a  given  point  to  drop  a  perpendicular 
to  a  given  line. 

Construction.    Draw  two  equal  line  seg-      v 
ments  from  the  given  point  to  the  given 
line  and  bisect  the  angle  between  them. 

Prove. 


48.  Problem  IV.     Construct   a   perpen- 
dicular bisector  to  a  straight   line-segment. 

Construct  as  in  the  adjacent  figure. 
Prove. 


\ 


49.  Problem  V.     At  a  given  point  on  a  given  straight  line 
to  construct  an  angle  equal  to  a  given  angle. 


Given  an  angle  ABC,  and  the  point  D  on  the  indefinite 
line  MAT. 

At  point  D  to  construct  an  angle  equal  to  angle  ABC. 

Analysis.     Suppose  the  work  done  and  that  angle  NDK 
is  equal  to  angle  ABC. 


Now  since  points  A}  C,  N,  K  can  have  any  position  on 
the  respective  arms  of  the  angles,  there  is  no  reason  why  BC, 
BA,  DN,  and  DK  should  not  be  made  the  same  length.  In 


II,  §  so]  TRIANGLES  39 

this  event  A  ACS  and  N KD  would  be  congruent,  and  thus 
N K  equal  AC.  Prove  this.  So  again  our  problem  resolves 
itself  into  the  construction  of  two  triangles  which  have  the 
three  sides  of  the  one  equal  to  the  three  sides  of  the  other. 

Construction.     Make  the  construction  and  give  the  proof. 

Problem  VI.  To  construct  a  triangle  having  given  two  sides 
and  the  included  angle. 

When  a  problem  is  merely  stated,  the  student  is  expected 
to  supply  figure,  analysis,  construction  and  proof. 

50.  Problem  VII.     Construct  a  triangle,  having  given  two 
angles  and  the  included  side. 

Problems. 

1.  Construct  triangles  whose  sides  are  as  follows: 

(a)  4  in.,  2.5  in.,  3  in.  (b)   2  in.,  5  in.,  4  in. 

(c)    10  cm.,  14  cm.,  10  cm.       (d)    15  cm.,  6  cm.,  10  cm. 

2.  (a)  Construct  the  triangle  whose  sides  are  3  in.,  3.5  in., 
4  in.,  respectively. 

(b)  Erect  perpendiculars  at  the  middle  points  of  the 
sides  of  this  triangle.    What  do  you  observe? 

(c)  Draw  the  bisectors  of  the  angles  of  this  triangle. 
What  do  you  observe? 

3.  Draw  any  triangle  and  repeat  (b)  and  (c)  of  Exercise  2. 

4.  Construct  a  quadrilateral  with  adjacent  sides  3  in.  and 
4.5  in.,  respectively,  with  opposite  sides  equal,  and  with  the 
long  diagonal  equal  to  6  inches. 

51.  Definition.     If  one  of  the  sides  of  a  triangle  is  pro- 
duced, the  angle  between  the 

side  produced  and  the  follow- 
ing side  is  called  an  exterior 
angle  of  the  triangle. 

Angle  MCA  is  an  exterior 
angle  of  triangle  A  BC.  Extend    A^~ 
other  sides  and  name  exterior  angles  thus  formed. 


40  PLANE  GEOMETRY  [H,  §  52 

52.  Theorem  XI.     The  exterior  angle  of  a  triangle  is  greater 
than  either  of  the  interior  angles  not  adjacent  to  it. 


We  have  given  the  triangle  ABC  with  the  side  AB  ex- 
tended through  the  vertex  By  forming  the  exterior  angle  XBC. 

We  are  to  prove  that  the  exterior  angle  XBC  is  greater  than 
either  of  the  interior  non-adjacent  angles  BAC  or  ACB. 

Analysis.  To  do  this  we  form  congruent  triangles.  We 
bisect  the  side  BC  and  call  the  mid-point  M.  Draw  a  line  from 
A  to  M.  Produce  AM  and  on  it  mark  the  point  P  so  that  MP 
shall  equal  AM.  Join  point  B  to  point  P  by  a  straight  line. 


This  gives  us  A  AMC  and  BPM  which  we  can  prove 
congruent  and  thus  get  a  comparison  of  Z  P£Cand  Z  ACB. 
By  comparing  Z  XBC  with  Z  PBC,  we  can  get  its  com- 
parison with  Z  ACB. 

Proof.     A  AMC  and  A  BPM  are  congruent.     Prove  this. 

How  does  Z  PBC  compare  with  Z  ACB?  Why? 

How  does  Z  XBC  compare  with  Z  PBC?  Why? 

Then  how  does  Z  XBC  compare  with  Z  ACB?     Why? 

By  producing  side  CB  through  vertex  B  prove  the  exterior 
angle  so  formed  greater  than  angle  A  and  thus  prove  the  prop- 
osition. 

State  the  proposition  proved. 


II,  §  53] 


TRIANGLES 


41 


53.  Theorem  XII.  //  two  sides  of  a  triangle  are  unequal, 
the  angles  opposite  these  sides  are  unequal,  and  the  longer  side 
has  the  greater  angle  opposite  it. 


Given  the  triangle  ABC,  with  side  b  longer  than  side  a. 
We  are  to  prove  that  angle  B  is  greater  than  angle  A. 

Analysis.  What  have  we  had  about  angles  opposite  sides 
of  triangles?  (See  Theorem  VII.)  On  side  b  lay  off  a  dis- 
tance CD  equal  to  side  a.  This  gives  us  two  angles  BDC 
and  CBD  that  we  can  prove  to  be  equal. 


We  can  now  compare  Z  BAG  with  Z  BDC,  and  Z  CBA 
with  Z  CBD,  and  thus  get  a  comparison  of  A  BAC  and 
CBA. 

Proof.  Name  the  isosceles  triangle  formed  when  line  BD 
is  drawn.  Why  is  it  isosceles? 

Name  the  angles  in  this  triangle  which  are  equal.  Why 
equal? 

What  is  Z  BDC  with  regard  to  the  A  ABDt 

How  does  it  compare  with  Z  A?     Why? 

How  does  Z  CBA  compare  with  Z  CBD?    Why? 

Therefore  how  does  Z  CBA  compare  with  Z  BAC?   Why? 

Assuming  that  side  c  is  greater  than  side  b,  what  is  true 
about  the  angles  opposite  these  sides. 

State  the  proposition  you  have  proved. 


42  PLANE  GEOMETRY  [II,  §  54 

Exercises: 

1.  The  interior  angles  of  a  triangle  are  40°,  60°,  and  80°, 
respectively.     Find  the  value  of  each  of  the  exterior  angles. 

2.  If  one  of  the  equal  angles  of  an  isosceles  triangle  is  35°, 
and  the  sum  of  the  three  angles  is  180°,  find  the  value  of 
each  exterior  angle  of  the  triangle. 

3.  Construct  a  triangle  whose  sides  shall  be  5,  6,  and  7 
units  long,  respectively.     Which  angle  is  the  largest? 

4.  If  two  oblique  lines  are  drawn  from  a  given  point  to  a 
given  line,  the  longer  oblique  line  makes  the  lesser  angle 
with  the  given  line. 

PART   III— TRANSVERSALS.        PARALLEL   LINES. 

SUMS  OF  ANGLES  OF  POLYGONS. 

PARALLELOGRAMS. 

54.  Definitions.  Draw  two  straight  lines  AB  and  CD. 
Cross  them  with  a  third  straight  line  EH.  Let  the  points 
of  intersection  be  F  and  G. 


The  angles  GFB  and  FGC  are  called  alternate  interior 
angles. 

The  angles  BFE  and  CGH  are  called  alternate  exterior 
angles. 

The  angles  BFE  and  DGF  are  called  corresponding  (or 
exterior-interior)  angles. 

Name  other  sets  of  alternate  interior  angles,  alternate 
exterior  angles,  and  corresponding  angles. 


II,  §  55]  PARALLELS  AND  TRANSVERSALS  43 

55.  Theorem  XIII.     //  two  straight  lines  are  cut  by  a  trans- 
versal making  one  set  of  alternate  interior  angles  equal,  then 
(a)   the  other  set  of  alternate  interior  angles  are  equal. 


To  construct  this  figure,  first  draw  two  intersecting  lines 
AB  and  EH  cutting  at  point  F.  Mark  any  other  point  on 
line  EH.  Call  it  G.  At  point  G,  using  compass  and  straight 
edge,  make  the  alternate  interior  angle  to  Z  GFB  equal  to  it. 
(See  Problem  V,  Article  49.) 

Given  the  straight  lines  AB  and  CD  cut  by  the  transversal 
EH  at  points  F  and  G,  making  the  angles  GFB  and  FGC 
equal. 

We  are  to  prove  that 

(a)  the  alternate  interior  A  DGF  and  AFG  are  equal. 

Analysis.  In  order  to  prove  that  Z  DGF  equals  Z  AFG 
we  shall  show  that  they  are  the  supplements  of  equal  angles, 
that  is  we  shall  show  that  Z  DGF  is  the  supplement  of 
Z  FGC,  and  that  Z  AFG  is  the  supplement  of  Z  GBF.  This 
will  be  sufficient  since  we  made  Z  FGC  equal  to  Z  GFB. 

Proof.     Z  DGF  is  the  supplement  of  Z  FGC.    Why? 

Z  AFG  is  the  supplement  of  Z  GFB.    Why? 

Z  FGC  is  equal  to  the  Z  GFB.  Why? 

Therefore  Z  DGF  equals  Z  AFG.          Why? 

State  the  theorem  proved. 

In  like  manner  let  the  student  give  analysis  and  proof  of 
the  following  statements. 

(6)    the  sets  of  alternate  exterior  angles  are  equal; 


44  PLANE  GEOMETRY  [II,  §  56 

(c)  the  sets  of  corresponding  angles  are  equal; 

(d)  the  sum  of  the  interior  angles  on  the  same  side  of  the 
transversal  is  equal  to  a  straight  angle; 

(e)  the  sum  of  exterior  angles  on  the  same  side  of  the  trans- 
versal is  equal  to  a  straight  angle. 

Corollary.  //  two  straight  lines  are  cut  by  a  transversal, 
making  a  set  of  alternate  exterior  angles  equal,  the  other  sets 
of  angles  of  the  figure  are  equal. 

There  are  three  more  statements  of  this  kind  which  the 
student  should  make  and  consider. 

Exercises: 

1.  Two  lines  are  crossed  by  a  transversal  so  that  two  of  the 
alternate  interior  angles  are  40°  and  70°  respectively.     Find 
the  value  of  each  angle  of  the  figure. 

2.  In  a  figure  formed  by  two  lines  crossed  by  a  transversal, 
the  sum  of  one  pair  of  alternate  interior  angles  is  150°.     What 
is  the  sum  of  the  other  pair? 

56.  Definition.    If  two  straight  lines  in  a  plane  do  not 
meet  however  far  they  are  produced,  they  are  said  to  be 
parallel. 

57.  Postulate  VI.    Postulate  of  Parallels.    The  following 
truth  is  assumed  in  geometry: 

Two  intersecting  lines  cannot  both  be  parallel  to  the  same 
straight  line. 

Historical  Note.  This  assumption,  or  its  equivalent  as  given  by 
Euclid,  has  been  much  discussed  by  mathematicians,  especially  during 
the  past  three  centuries.  Euclid  assumed  that,  when  two  lines  AB  and 
CD  are  crossed  by  a  transversal  (see  figure  on  page  42),  if  the  sum  of 
the  interior  angles  on  one  side  of  the  transversal  is  less  than  a  straight 
angle,  the  lines  AB  and  CD,  if  produced,  will  meet  on  that  side  of  the 
transversal. 

This  was  long  suspected  to  be  a  theorem,  capable  of  proof,  and  many 
attempts  were  made  to  prove  it,  but  without  success.  Euclid  no  doubt 
made  such  attempts  and  his  keen  insight  is  shown  by  the  fact  that  he 
finally  stated  it  outright  as  a  thing  which  must  be  assumed. 


n,  §  58]  PARALLELS  AND  TRANSVERSALS  45 

58.  Theorem  XTV.  //  two  straight  lines  are  cut  by  a  trans- 
versal making  a  set  of  alternate  interior  angles  equal,  the  lines 
are  parallel. 


Given  the  two  straight  lines  LK  and  SR  cut  by  the  trans- 
versal TH  in  the  points  D  and  G  respectively,  making  the 
angle  DOS  equal  to  the  angle  GDK.  (Use  compass  to  make 
the  angles  equal.) 

We  are  to  prove  that  lines  LK  and  SR  are  parallel. 

Analysis.  In  order  to  prove  that  lines  LK  and  SR  are 
parallel,  we  must  prove  that  they  do  not  meet  no  matter 
how  far  they  are  produced.  To  prove  this  we  shall  suppose 
that  they  do  meet  in  some  distant  point,  say  point  P  lying 
to  the  right  of  TH,  which  is  too  far  away  to  be  marked  on 
the  page,  thus  forming  A  GPD  whose  exterior  angle  is  DGS. 
If  this  were  true  we  should  have  Z  DGS  both  equal  to 
Z  GDK  and  greater  than  Z  GDK  at  the  same  time,  which 
would  be  impossible.  Therefore  the  lines  can  not  meet, 
since  this  would  produce  an  impossible  condition. 

Proof.  Suppose  that  the  lines  LK  and  SR  meet  at  a 
distant  point  P  to  the  right. 

Name  the  triangle  that  will  be  formed. 

What  is  Z  DGS  with  reference  to  this  triangle? 

What  then  is  the  comparison  of  A  DGS  and  GD  Kt    Why? 

By  hypothesis  what  is  the  comparison  of  A  DGS  and 
GDKt 

What  must  be  the  conclusion  from  these  two  statements? 


46  PLANE  GEOMETRY  [II,  §  59 

Suppose  that  the  lines  meet  at  some  distant  point  to  the 
left,  say  Q. 

By  reasoning  similar  to  that  just  given  show  that  this  is 
impossible. 

State  the  proposition  proved. 

Corollary.  //  two  straight  lines  are  cut  by  a  transversal, 
making  a  set  of  alternate  exterior  angles  equal,  the  lines  are 
parallel. 

State  other  theorems  of  this  kind  using  the  other  sets  of 
angles. 

Problem  VIII.  Through  a  given  point  to  draw  a  line 
parallel  to  a  given  line. 

This  depends  directly  on  Theorem  XIV. 

59.  Theorem  XV.  //  two  parallel  straight  lines  are  cut  by 
a  transversal  (a)  the  alternate  angles  are  equal,  (b)  the  cor- 
responding angles  are  equal,  (c)  the  sum  of  the  interior  angles 
on  the  same  side  of  the  transversal  is  a  straight  angle,  and  (d) 
the  sum  of  the  exterior  angles  on  the  same  side  of  the  transversal 
is  a  straight  angle. 


Given  the  parallel  lines  KL  and  NM  cut  by  the  trans- 
versal TR  in  the  points  S  and  D  respectively. 

We  are  to  prove  that  angle  MDS  equals  angle  KSD,  etc. 

Analysis.  To  prove  that  Z  MDS  is  equal  to  Z  KSD,  we 
take  a  different  plan  from  any  which  we  have  given  before. 
Evidently  one  of  the  following  statements  must  be  true: 


n,  §  59]  PARALLELS  AND  TRANSVERSALS  47 

either  Z  MDS  is  less  than  Z  KSD, 

or  Z  MD£  is  greater  than  Z 

or  Z  MZ)£  is  equal  to  Z 

If  we  can  prove  that  two  of  these  statements  are  false,  the 

third  must  be  true. 

Taking  up  the  first  statement  we  suppose  that  Z  MDS 
is  less  than  Z  KSD.  If  this  is  true,  in  Z  KSD  we  can  lay 
off  an  Z  ISD  equal  to  Z  MDS,  in  which  case  the  line  IS  will 
lie  within  Z  X/SD  as  shown  in  the  figure  below.  Now  since 

T 


M 


A  MDS  and  ISD  are  equal,  line  IS  is  parallel  to  line  NM. 
So  both  the  intersecting  lines  KL  and  7$  are  parallel  to 
line  NM.  Therefore  the  assumption  that  Z  MDS  is  less 
than  Z  X$D  leads  to  an  impossibility,  and  we  conclude  that 
Z  MDS  is  not  less  than  an  Z  KDS. 

By  a  similar  line  of  proof  we  can  show  that  the  second 
statement  is  false  thus  leaving  the  third  one  true. 

Proof.    Suppose  Z  MDS  <  Z  KSD}  and  Z  ISD  =  Z  MDS. 

The  line  IS  falls  within  Z  XSD.    Why? 
/.  IS  ||  Mlf.  Why? 

But  KL  ||  JVM.  Why? 

Therefore  two  intersecting  lines  KL  and  IS  are  parallel 
to  the  same  line  NM,  which  is  impossible.  Why?  (See 
postulate  of  parallels  Art.  57.) 

Therefore  Z  MDS  cannot  be  less  than  Z  KSD.  In  like 
manner  prove  that  Z  MDS  cannot  be  greater  than  Z  KSD. 

This  proves  that  Z  MDS  =  Z  #SD. 

Prove  the  other  statements  by  use  of  Theorem  XIII. 


48  PLANE  GEOMETRY  [ii,  §  60 

60.  Theorem  XVI.     A  line  perpendicular  to  one  of  two 
parallel  lines  is  perpendicular  to  the  other. 

Prove  this  by  making  use  of  Theorem  XV  and  the  definition 
of  perpendicular  lines. 

61.  Theorem  XVII.     A  line  parallel  to  one  of  two  parallel 
lines  is  parallel  to  the  other. 

Cross  the  three  lines  by  a  transversal,  and  apply  Theorems 
XV  and  XIV. 

62.  Theorem  XVIII.     Two  angles  whose  arms  are  parallel 
each  to  each  are  either  equal  or  supplementary. 

They  are  equal  when  the  corresponding  arms  extend  in 
the  same  direction  from  the  vertex  or  in  opposite  directions 
from  the  vertex,  and  supplementary  when  one  pair  of  cor- 
responding arms  extends  in  the  same  direction  and  the  other 
in  the  opposite  direction. 


\ 


Give  this  proof  by  extending  the  arms  until  they  intersect, 
and  applying  Theorem  XV  and  the  axiom — Things  equal  to 
the  same  thing  are  equal  to  each  other. 

Exercise : 

1.  Draw  an  angle  ABC,  and  draw  two  intersecting  lines, 
one  parallel  to  each  arm  of  the  angle.     Produce  these  lines 
to  cut  the  arms  of  the  angle,  forming  a  quadrilateral.     Com- 
pare the  various  angles  of  the  figure,  both  interior  and  ex- 
terior, with  angle  ABC. 

2.  Show  how  a  triangle,  cut  from  cardboard,  can  be  used 
to  draw  lines  parallel  to  a  given  line. 


II,  §  63] 


PARALLELS  AND  TRANSVERSALS 


49 


63.  Theorem    XIX.     Two    angles    whose    arms    are    per- 
pendicular each  to  each  are  either  equal  or  supplementary. 


A  P 


Give  this  proof  by  drawing,  through  the  vertex  of  one  of  the 
angles,  lines  parallel  to  the  arms  of  the  other  as  shown  in  the 
third  figure.  Apply  Theorems  XVIII  and  I. 

64.  Exercises  depending  on  Theorems  XIII  to  XIX 
inclusive. 

Read  again  Article  36;  (c)  will  now  read:  If  I  must  prove 
two  angles  equal,  can  I  prove  that  they  are 

alternate  interior  angles; 
or  alternate  exterior  angles; 

or  corresponding  angles  of  parallel  lines? 

If  I  must  prove  lines  parallel,  can  I  prove  that,  being  cut 
by  a  transversal, 

the  alternate  interior  angles 
or  the  alternate  exterior  angles 

or  the  corresponding  angles 

are  equal?  If  I  must  prove  these  equal,  can  I  prove  that  they 
are  the  corresponding  angles  of  congruent  triangles?  Or  can 
I  prove  them  equal  to  the  same  angle? 

1.  In  figure  under  Theorem  XV,  what  is  the  number  of 
degrees  in  each  of  the  angles  if  Z  DSL  is  20  degrees?  108 
degrees?  a  degrees?  (a  +  6)  degrees?  a(a  +  36)  degrees? 


50  PLANE  GEOMETRY  [II,  §  64 

2.  In  the  same  figure  how  many  radians  in  each  of  the 
angles  of  the  figure  if  Z  RDM  is  \  x  radians?     f  IT  radians? 
a  +  ^  TT  radians?     2(r  +  j)  TT  radians?     (?  -  p  TT  radians? 
1J  radians?     a  +  r-  radians?     Express  each  answer  as  an 
improper  fraction  when  possible. 

3.  Find  the  number  of  degrees  in  each  angle  of  the  same 
figure,  if  Z   TSL  is  2J  degrees  more  than  4  times  Z  NDS. 

4.  If  Z  SDN  is  •§•  Z  XSD,  find  the  number  of  radians  in 
each  angle  of  the  same  figure. 

5.  If  Z  SDN  is  I  part  of  Z   #SD,  find  the  number  of 
radians  of  each  angle  of  the  figure. 

6.  If  Z  LST  is  ]  degrees  more  than  r  times  Z  RDM,  how 
many  degrees  are  there  in  each  of  the  angles  of  the  figure? 

7.  Write  three  problems  involving  Theorems  XIII,  XV. 

8.  Two  lines  AB  and  CD  are  cut  by  a  transversal  DE  in 
the  points  F  and  G  respectively,  forming  the  alternate  in- 
terior angles  GFB  and  FGC.     Z  GFB  is  8  degrees  more  than 
2  times  Z  FGC.     If  Z  GFB  were  made  28  degrees  less  and 
Z  FGC  were  made  40  degrees  more  the  lines  would  be  par- 
allel.    Find  the  number  of  degrees  in  each  angle  as  they  are 
drawn. 

9.  Two  lines  are  crossed  by  a  transversal  making  one  of 
the  two  alternate  exterior  angles  equal  to  the  ™  part  of  the 
other.     If  J?  part  of  a  right  angle  is  added  to  the  one  and 
£  part  of  a  right  angle  is  subtracted  from  the  other,  the  lines 
will   be   parallel.      Find   the   number   of   degrees   in   each 
angle. 

10.  If  PQR  is  an  isosceles  triangle  with  side  p  equal  to 
side  q,  the  bisector  of  the  exterior  angle  formed  at  R  by  pro- 
ducing side  q  is  parallel  to  side  PQ. 

11.  If  a  point  is  placed  between  two  parallel  lines,  and 
through  the  point  two  intersecting  lines  are  drawn  cutting 
the  parallel  lines,  two  mutually  equiangular  triangles  are 
formed. 


II,  §  64]  PARALLELS  AND  TRANSVERSALS  51 

12.  If  two  parallel  lines  are  cut  by  a  transversal,  and  each 
angle  of  a  pair  of  alternate  interior  angles  is  bisected,  the 
bisectors  are  parallel. 

13.  If  two  parallel  lines  are  cut  by  a  transversal,  and  the 
interior  angles  on  the  same  side  of  the  transversal  are  bisected, 
the  bisectors  are  perpendicular  to  each  other. 

14.  Cut  a  triangle  from  stiff  paper.    By  sliding  it  along  a 
straight  line  show  how  it  may  be  used  to  draw  parallel  lines. 

(A  right-angled  triangle  is  commonly  used  for  this  purpose.) 

15.  If  through  any  point  on  the  hypotenuse  of  a  right- 
angled  triangle  a  line  is  drawn  parallel  to  one  of  the  sides,  it 
is  perpendicular  to  the  other. 

16.  If  through  any  point  on  the  hypotenuse  of  a  right- 
angled  triangle,  a  line  is  drawn  perpendicular  to  one  of  the 
sides  it  is  parallel  to  the  other. 

17.  M  is  any  point  on  the  hypotenuse  of  a  right  triangle 
ABC  of  which  CA  is  the  hypotenuse.     Through  M  two  lines 
are  drawn,  one  parallel  to  side  AB  cutting  BC  in  point  R, 
and  one  perpendicular  to  CA  cutting  BC  (or  BC  produced)  in 
S.     Prove  that  A  ABC  and  MSR  are  mutually  equiangular. 

18.  In  the  figure  AB  \\  CD,  AR  \\ 
BD,  and  BS  \\  AC.      Prove  that 
A  CRA  and  SDB  are  mutually 
equiangular. 

19.  Construct  a  triangle,  equi- 
angular with  a  given  triangle,  by  drawing  lines  parallel  to 
the  sides  of  the  given  triangle.     Prove. 

20.  Bisect  the  four  angles  of  a  quadrilateral  whose  opposite 
sides  are  parallel,  and  produce  the  bisectors  to  intersect,  forming 
a  new  quadrilateral.     Show  that  all  its  angles  are  right  angles. 

21.  In  a  quadrilateral  with  opposite  sides  parallel,  a  diagonal 
bisects  a  pair  of  opposite  angles.     Prove  the  four  sides  equal. 

22.  Construct  an  angle  of  60°. 

Suggestion.     Construct  an  equilateral  triangle,  and  assume 
that  the  sum  of  the  three  angles  is  180°. 


52 


PLANE  GEOMETRY 


[II,  §  64 


\A/y\AAA 

vvvvV 


23.  Construct  two  lines  intersecting  at  an  angle  of  60°. 
Call  one  AB  and  the  other  CD.  Regarding  AB  as  a  trans- 
versal lay  off  a  number  of  equal  segments  on  it.  Through 
the  points  of  division 
construct  lines  parallel 
to  CD.  Regarding  CD 
as  a  transversal  lay  off 
a  number  of  segments 
equal  to  those  on  AB. 
Through  the  points  of 
division  construct  lines 
parallel  to  AB.  By  drawing  lines  through  the  points  of 
intersection  you  will  have  a  design  of  equilateral  triangles. 


24.  Construct  this  design, 
formed  of  equilateral  trian- 
gles. It  may  be  used  as  pat- 
tern for  making  a  model  of 
one  of  the  so-called  regular 
solids,  namely,  the  regular 
octahedron. 


25.  By  constructing  parallel  lines  and 
darkening  portions  bring  out  the  adja- 
cent design. 


26.    As   in  Exercise  25,  for  this  tiling 
pattern. 


27.  Make  a  design  of  your  own,  based  on  equilateral  tri- 
angles. 


II,  §  64]  PARALLELS  AND  TRANSVERSALS 

28.  Outline  the  following  designs. 


53 


29.  By  constructing  parallel  lines  and 
darkening  portions  bring  out  the  ad- 
jacent design. 


30.  Make  drawing  showing  plan  of  a  stair- 
way having  four  steps  to  the  first  landing, 
then  four  steps  to  the  second  landing,  then 
six  steps  to  the  top. 

31.  To  divide  a  board  lengthwise  into  a  given  number  of 
equal  parts;  for  example,  to  divide  a  6-inch  board  into  5 
equal  parts. 


Place  the  square  or  ruler  across  the  board  as  shown  in  the 
figure,  so  that  the  two-inch  and  the  twelve-inch  mark  shall 
fall  on  the  edges  of  the  board.  Mark  on  the  board  at  the 
readings  4,  6,  8,  10.  Move  the  square  to  a  second  position, 
as  in  (2).  Again  mark  the  points  4,  6,  8,  10.  Join  the 
marks  with  corresponding  numbers.  Explain  why. 


54  PLANE  GEOMETRY  [n,  §  65 

65.  Theorem  XX.     The  sum  of  the  angles  of  a  triangle  is 
equal  to  a  straight  angle. 


Given  the  triangle  ABC. 

To  prove  that  Z4  +  Z5+ZC=a  straight  angle. 

Analysis.  To  do  this,  we  shall  construct  three  angles 
which  shall  equal  respectively  the  angles  of  the  triangle,  and 
whose  sum  shall  equal  a  straight  angle.  To  do  this  produce 
the  side  AB  through  B  to  X,  as  in  the  figure.  Through 
B  draw  a  line  BY  parallel  to  side  AC.  This  gives  three 
angles  whose  sum  is  a  straight  angle.  If  we  can  prove 
A  XBY,  YBC,  and  CBA  equal  respectively  to  A  BAG, 
ACB,  and  CBA,  we  shall  have  proved  our  theorem. 

Proof.  Z  XBY  =  Z  BAG.  Why? 
Z  YBC  =  Z  ACB.  Why? 
Z  CBA  =  Z  CBA.  Why? 

Z  XBY  +  Z    YBC  +    Z   CBA    =  a   straight 

angle. 

Z  BAG  +  Z    AC£  +    Z    CBA    =   a  straight 

angle.  Why? 

State  the  theorem  proved. 

Corollary  1.  The  exterior  angle  of  a  triangle  is  equal  to  the 
sum  of  the  interior  angles  not  adjacent  to  it. 

Corollary  2.  If  a  triangle  has  one  right  angle  or  one  obtuse 
angle,  each  of  the  other  angles  is  acute. 


II,  §  66]  PARALLELS  AND  TRANSVERSALS  55 

Corollary  3.  //  two  angles  of  a  triangle  are  equal  respectively 
to  two  angles  of  another  triangle,  the  third  angles  are  equal. 

Corollary  4.  //  two  triangles  have  two  angles  and  any  side 
of  one  equal  respectively  to  two  angles  and  the  corresponding 
side  of  the  other,  the  triangles  are  congruent. 

Corollary  5.  //  two  right  triangles  have  an  acute  angle  and 
a  side  of  one  equal  respectively  to  an  acute  angle  and  the  cor- 
responding side  of  the  other,  the  triangles  are  congruent. 

66.  Theorem  XXI.     //  a  triangle  has  two  of  its  angles  equal, 
the  sides  opposite  are  equal  and  the  triangle  is  isosceles. 

To  prove  this  divide  the  triangle  into  two  triangles  by 
drawing  a  line  from  the  vertex  of  the  third  angle  perpendicular 
to  the  side  included  between  the  equal  angles. 

67.  Problem  IX.    Given  two  angles  of  a  triangle,  to  con- 
struct the  third  angle. 

68.  Problem  X.    Given  two  angles  of  a  triangle  and  the  side 
opposite  one  of  them,  to  construct  the  triangle. 

69.  Problem  XI.    Given  two  sides  of  a  triangle  and  the 
angle  opposite  one  of  them,  to  construct  the  triangle. 


Given  the  angle  A  and  the  sides  a  and  6  of  a  triangle,  the 
side  a  being  opposite  angle  A. 

To  construct  the  triangle. 

Construction.     Draw  XAY  equal    to  the  given  angle. 
On  AY  lay  off  AC  equal  to  6. 

With  C  as  a  center  and  a  radius  a  draw  arc 


56  PLANE  GEOMETRY  pi,  §  70 

Then  the  first  figure  shows  two  possible  triangles,  ABC 
and  AB&. 

The  second  figure,  where  a  is  longer  than  6,  shows  only  one 
possible  triangle.  Give  proof  in  each  case. 

Let  the  student  draw  figures  with  shorter  values  of  a. 
How  long  must  a  be  to  make  the  construction  possible? 

Study  also  the  cases  where  angle  A  is  an  obtuse  angle. 

70.  Theorem  XXII.  //  two  angles  of  a  triangle  are  unequal, 
the  sides  opposite  are  unequal  and  the  greater  angle  has  the 
longer  side  opposite  it. 


Given  the  triangle  ABC  with  the  angle  B  greater  than 
angle  A. 

To  prove  that  side  b  is  longer  than  side  a. 

Analysis.  Since  we  now  know  that  if  two  angles  of  a 
triangle  are  equal  the  sides  opposite  are  equal,  we  draw  in 
the  line  EM  making  Z  MBA  equal  to  Z  BAG.  Line  BM 
will  fall  between  sides  c  and  a,  because  Z  CBA  is  greater 
than  Z  A.  Now  line  BM  equals  AM',  also  CM  +  MB  is 
greater  than  CB,  since  a  straight  line  is  the  shortest  distance 
between  two  points.  So  we  have  proved  the  theorem  when 
we  have  proved  the  statements  made  in  our  analysis. 

Proof.    BM  =  MA.  Why? 

CM  +  MB  >  a.  Why? 

CM  +  MA  >  a.  Why? 

6  >  a.  Why? 
State  the  proposition  proved. 


II,  §  71]  PARALLELS  AND  TRANSVERSALS  57 

Corollary  1.  The  hypotenuse  of  a  right-angled  triangle  is 
the  longest  side  of  the  triangle. 

Corollary  2.  Of  all  the  lines  drawn  from  an  external  point  to 
a  given  straight  line  the  perpendicular  is  the  shortest. 

Corollary  3.  Of  two  oblique  lines  drawn  from  an  external 
point  to  a  given  straight  line,  the  one  which  cuts  the  greater 
distance  from  the  foot  of  the  perpendicular  is  the  longer. 

Suggestion.  In  the  figure  let 
PH  be  the  perpendicular  from 
PonlineZF.  Z.  TRP  is  ob- 
tuse because  it  is  an  exterior 
angle  to  A  HRP.  Hence  PT 
is  longer  than  PR,  since  it 
lies  opposite  the  greatest  angle 
in  A  RTP.  x — 

71.  In  the  following  exercises  observe  the  suggestions 
below: 

If,  the  relations  of  angles  being  given,  it  is  required  to 
find  the  values  of  the  angles,  form  an  equation  by  applying 
one  of  the  following: 

Theorem.    The  sum  of  the  angles  of  a  triangle 

equals  a  straight  angle. 
Or  apply         Corollary.     The  sum  of  the  acute  angles  of  a 

right-angled  triangle  equals  a  right  angle. 
Or  apply        Axiom.     Things  equal  to  the  same  thing  are 

equal  to  each  other. 

For  examples  and  illustrative  solutions,  see  First  Course,  pages  37  and  38,  Exercises 
9  to  21. 

To  prove  two  lines  equal,  prove  that  they  are  corresponding 
sides  of  congruent  triangles;  or  prove  that  they  are  opposite 
the  equal  angles  of  an  isosceles  triangle. 

To  prove  that  a  triangle  is  isosceles,  prove  that  two  of  its 
angles  are  equal,  or  that  two  of  its  sides  are  equal. 


58  PLANE  GEOMETRY  [II,  §  72 

To  prove  that  two  angles  are  equal,  prove 
that  they  are  the  alternate  interior  angles; 
or  they  are  the  alternate  exterior  angles; 
or  they  are  the  corresponding  angles  of  parallel  lines; 
or  they  have  their  arms  parallel  each  to  each; 
or  they  have  their  arms  perpendicular  each  to  each; 
or  they  are  vertical  angles; 
or  they  are  complements  of  equal  angles; 
or  they  are  equal  to  equal  angles; 
or  they  are  halves  of  equal  angles; 
or  they  are  corresponding  angles  of  congruent  triangles. 
To  prove  two  triangles  congruent,  prove  that 

they  have  two  sides  and  the  included  angle  of  the  one 

equal  respectively  to  two  sides  and  the  included  angle 

of  the  other. 
or  they  have  two  angles  and  the  included  side  of  the  one 

equal  to  two  angles  and  the  included  side  of  the 

other. 
or  they  have  three  sides  of  the  one  equal  respectively  to 

the  three  sides  of  the  other. 
or  they  have  two  angles  and  an  opposite  side  of  the  one 

equal  to  two  angles  and  an  opposite  side  of  the 

other. 
To  prove  two  triangles  mutually  equiangular,  prove  that 

they  have  two  angles  of  one  equal  respectively  to  two 

angles  of  the  other;  then  the  third  angles  are  equal. 

72.  Exercises. 

1.  Prove  that  the  sum  of  the  angles  of  a  triangle  is  a 

straight  angle  by  using  the  ad-  c - 

j  acent  figure,  in  which  the  line  M  ....------ 

MN  is  drawn  parallel  to  the  side 
AB. 

2.  The  sum  of  the  acute  angles 
of  a  right  triangle  is  i  TT  radians. 


II,  §  72]  PARALLELS  AND  TRANSVERSALS  59 

3.  If  one  angle  of  a  right  triangle  is  45°,  the  triangle  is 
isosceles. 

4.  If  one  acute  angle  of  a  right  triangle  is  twice  the  other, 
find  the  number  of  degrees  in  each  angle  of  the  triangle. 
Show  that  such  a  triangle  may  be  formed  by  bisecting  an 
equilateral  triangle. 

5.  In  the  triangle  of  Exercise  4  prove  that  the  hypotenuse 
is  twice  the  side  opposite  smaller  angle. 

Suggestion.  Draw  a  line  through  the  vertex  of  the  right 
angle  cutting  the  hypotenuse  in  such  a  way  that  two  isosceles 
triangles  will  be  formed.  Or,  bisect  an  angle  of  an  equilateral 
triangle. 

6.  If  the  largest  of  the  three  angles  of  a  triangle  is  31 
degrees  more  than  3  times  the  smallest,  and  33  degrees  more 
than  2  times  the  intermediate  angle,  what  is  the  number  of 
degrees  in  each  angle  of  the  triangle? 

7.  If  the  first  angle  of  a  triangle  is  30°  more  than  twice  the 
second,  and  the  third  is  12°  more  than  the  sum  of  the  first 
and  second,  what  is  the  number  of  degrees  in  each  angle 
of  the  triangle,  both  interior  and  exterior? 

8.  One  of  two  angles  of  a  triangle  is   15  f  degrees  less 
than  §  of  the  other.     If  the  side  included  by  these  angles 
be  turned  so  that  the  smaller  angle  becomes  37  degrees  more 
and  the  larger  angle  is  37  degrees  less,  the  triangle  will  be 
isosceles.     Find  the  number  of  degrees  in  each  angle  of  the 
triangle. 

9.  In  a  right  triangle  one  of  the  acute  angles  is  6°  less  than 
|  of  the  other.     How  many  degrees  must  be  taken  from  the 
larger  and  added  to  the  smaller  angle  to  make  the  hypotenuse 
twice  the  shortest  side  of  the  triangle. 

10.  One  of  the  angles  of  an  isosceles  triangle  is  £  TT  radians; 
what  is  the  number  of  radians  in  each  angle  of  the  triangle 
both  interior  and  exterior. 


60  PLANE  GEOMETRY  [II,  §  72 

11.  If  one  of  the  angles  of  a  triangle  is  r  degrees  more  than 
a  times  another,  and  the  difference  between  the  two  is  | 
part  of  the  third,  find  the  number  of  degrees  in  each  angle 
of  the  triangle,  both  interior  and  exterior. 

12.  The  angle  between  the  equal  sides  of  an  isosceles  tri- 
angle is  m  degrees.     A  line  is  drawn  to  divide  it  into  two 
parts  such  that  p  times  the  greater  after  it  is  increased  by 
p  degrees  is  equal  to  r  times  the  less  after  it  is  diminished  by 
|  degrees.     Find  the  number  of  degrees  in  each  angle  of  the 
figure. 

13.  Write  a  problem  concerning  the  angles  of  a  triangle 
using  special  relations. 

14.  Write  a  problem  concerning  the  angles  of  a  triangle 
using  general  numbers. 

15.  If  the  angles  here  shown  are  two  angles  of  a  triangle, 
construct  the  third  angle. 


16.  If   A  A  and  B  are  the  angles  of  a  triangle,  and  line 
MN  is  the  side  opposite  Z  A  construct  the  triangle. 


II,  §  72]  PARALLELS  AND  TRANSVERSALS  61 

17.  If  through  a  point  on  the  bisector  of  an  angle  a  line 
is  drawn  parallel  to  one  arm  of  the  angle  and  cutting  the 
other,  an  isosceles  triangle  is  formed. 

18.  If  A  A  and  B  of  A  ABC  are  bi- 
sected by  lines  intersecting  in  P,  and  a 
line  HK  is  drawn  through  P  parallel  to 
AB,  the  line  HK  is  equal  to  the  sum  of 
the  segments  AH  and  BK. 

19.  If  from  any  point  in  the  bisector  of  an  angle  lines  are 
drawn  perpendicular  to  the  arms  of  the  angle  two  congruent 
triangles  are  formed,  and  hence  the  perpendiculars  are  equal. 
This  proposition  may  be  worded: 

Any  point  on  the  bisector  of  an  angle  is  equidistant  from  the 
arms  of  the  angle. 

20.  If  the  equal  angles  of  an  isosceles  triangle  are  bisected, 
and  the  bisectors  are  extended  to  intersect  the  opposite  sides, 
two  sets  of  congruent  triangles  are  formed. 

21.  The  mid-point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices. 

22.  If,  in  an  isosceles  triangle,  the  two  equal  angles  are  bi- 
sected, and  the  bisectors  are  extended  to  intersect,  a  new 
isosceles  triangle  is  formed. 

23.  In  Exercise  22  the  angle  at  the  vertex  of  the  first  triangle 
is  100°;  what  is  the  angle  at  the  vertex  of  the  new  triangle? 

24.  In  Exercise  22,  suppose  that  the  angle  at  the  vertex 
of  the  original  triangle  is  equal  to  one-half  of  one  of  the 
equal  angles.     Find  the  angles  of  the  original  triangle  and 
of  the  new  triangle. 

25.  In  Exercise  20,  suppose  each  of  the  equal  angles  to  be 
70°;  find  all  the  angles  formed  when  the  bisectors  are  drawn. 

26.  In  the  isosceles  triangle  PQR,  side  p  being  equal  to  side 
q,  the  side  q  is  produced  through  the  vertex  R  to  the  point  K 
so  that  RK  equals  q.     Line  QK  is  drawn.     Prove  that 
A  RQ  K  is  isosceles.     Prove  that  A  PQ  K  is  a  right  triangle. 


62  PLANE  GEOMETRY  [II,  §  73 

27.  If,  in  a  right  triangle,  a  line  is  drawn  from  the  vertex  of 
the  right  angle  perpendicular  to  the  hypotenuse,  two  triangles 
are  formed  which  are  mutually  equiangular  to  the  given  tri- 
angle and  to  each  other. 

28.  If  A  M  NO  is  isosceles  with  MO  =  NO,  and  a  line  is 
drawn  perpendicular  to  MN  cutting  M  N  in  R,  MO  in  Q 
and  NO  produced  in  P,  prove  that  A  QOP  is  isosceles. 

73.  Definitions. 

A  polygon  of  n  sides  is  in  general  called  an  n-gon. 
Special  names  are: 

3  sides:  triangle. 

4  sides:  quadrilateral. 

5  sides:  pentagon. 

6  sides:  hexagon. 

7  sides:  heptagon. 

8  sides:  octagon. 

9  sides:  nonagon. 
10  sides:  decagon. 
12  sides:  dodecagon. 

A  polygon  is  regular  when  its  sides  are  equal  and  its 
angles  are  equal. 

A  diagonal  of  a  polygon  is  a  line  drawn  from  one  vertex 
to  any  other  vertex  not  adjacent  to  it. 

74.  Theorem  XXIII.     The  sum  of  the  angles  of  an  n-gon  is 
equal  to  (n  —  2)  straight  angles. 

In  order  to  prove  this  we  will  examine  several  different 
polygons.  Draw  a  quadrilateral.  Draw  one  of  its  diag- 
onals. Into  how  many  triangles  does  the  diagonal  divide 
the  quadrilateral?  What  is  the  sum  of  the  angles  of  each 
triangle?  What  is  the  sum  of  the  angles  of  the  two  triangles 
and  hence  of  the  quadrilateral? 

Draw  a  pentagon.  Draw  the  diagonals  from  one  vertex. 
Into  how  many  triangles  do  the  diagonals  divide  the  pen- 


II,  §  75] 


PARALLELS  AND  TRANSVERSALS 


63 


tagon?  What  is  the  sum  of  the  angles  of  each  triangle? 
What  is  the  sum  of  the  angles  of  the  three  triangles  and 
hence  of  the  pentagon? 

Repeat  this  experiment  with  a  hexagon;  with  a  heptagon; 
with  a  nonagon. 

Make  a  table  of  your  answers  to  the  above,  thus: 


Number 
of  sides 

Number  of 
triangles 

Sum  of  angles 
in  one  triangle 

Sum  of  angles 
of  the  figure 

4 
5 

2=4-2 
3=5-2 

1  st.  angle 
1  st.  angle 

(4  —  2)  st.  angles 
(5  —  2)  st.  angles 

Do  you  find  that  you  can  tell  the  number  of  triangles  into 
which  a  20-angled  figure  can  be  divided  by  drawing  the 
diagonals  from  one  vertex?  A  38-angled  figure? 

What  is  the  sum  of  the  angles  of  the  20-angled  figure?  Of 
the  38-angled  figure? 

Into  how  many  triangles  will  the  diagonals  of  an  n-angled 
figure  divide  the  figure?  What  then  is  the  sum  of  the  angles 
of  an  n-angled  figure? 

Give  a  complete  proof  of  the  theorem. 

State  the  theorem  that  you  have  proved. 

75.  Theorem  XXIV.  The  sum  of  the  exterior  angles  of  a 
convex  polygon  is  equal  to  a  perigon. 


Given  the  polygon  ABC  ....  having  n  sides. 


64  PLANE  GEOMETRY  pi,  §  76 

Let  a  =  the  number  of  angular  units  in  the  interior  angle 

at  A, 
«i  =  the  number  of  angular  units  in  the  exterior  angle 

at  A, 
b  =  the  number  of  angular  units  in  the  interior  angle 

at  B, 
61  =  the  number  of  angular  units  in  the  exterior  angle 

at  B,  and  so  on. 

To  prove  that  d  +  61  +  Ci  +  etc.  =  a  perigon. 

Analysis.  To  prove  this  we  find  the  sum  of  the  interior 
and  exterior  angles  at  each  vertex,  and  then  by  multiplying 
by  the  number  of  vertices  we  shall  have  the  sum  of  the  in- 
terior and  exterior  angles  of  the  polygon.  Subtracting  the 
sum  of  the  interior  angles  we  arrive  at  the  sum  of  the  exterior 
angles. 

Proof.  What  is  the  value  of  a  +  ai? 
What  is  the  value  of  b  +  61? 
What  is  the  value  of  c  +  ci? 

Since  n  is  the  number  of  vertices  of  the  polygon,  the  sum  of 
the  interior  and  the  exterior  angles  is 

n  straight  angles.         Why? 
The  sum  of  the  interior  angles  is 

(n  —  2)  straight  angles.         Why? 

Then  the  sum  of  the  exterior  angles  is  the  amount  that 
must  be  added  to  (n  —  2)  straight  angles  to  get  n  straight 
angles,  or  it  is 

n  —  (n  —  2)  straight  angles, 
which  equals  2  straight  angles,  or  a  perigon. 
State  the  proposition  proved. 

76.  Exercises.    Angles  of  Polygons. 

1.  What  is  the  number  of  sides  of  a  polygon  if  the  sum 
of  its  interior  angles  is  equal  to  the  sum  of  its  exterior 
angles. 


II,  §  76]  PARALLELS  AND  TRANSVERSALS  65 

2.  Prove  Theorem  XXIII  by  placing  a  point  within  the 
polygon,  and  joining  it  to  the  vertices  of  the  polygon  as 
shown  in  the  first  figure. 


Again  prove  Theorem  XXIII  by  means  of  the  second 
figure. 

3*  What  is  the  number  of  degrees  in  each  angle  of  an 
equiangular  hexagon? 

4.  How  many  sides  has  a  polygon  the  sum  of  whose  in- 
terior angles  is  TT  radians? 

5.  In  a  pentagon  the  second  angle  is  5  degrees  larger  than 
the  first;  the  third  is  twice  the  size  of  the  second;  the  fourth 
is  twice  the  size  of  the  third;  the  fifth  is  two-thirds  the  size 
of  the  fourth.     What  is  the  number  of  degrees  in  each  angle 
of  the  pentagon? 

6.  In  a  pentagon  the  first  two  angles  are  equal;  the  third 
is  14  degrees  less  than  3  times  the  first;  the  fourth  is  4  times 
the  difference  between  the  third  and  second;  the  fifth  is  39 
degrees.     What  is  the  number  of  degrees  in  each  angle  of 
the  pentagon? 

7.  In  a  hexagon  the  second  angle  is  24°  more  than  the  first; 
the  third  is  24°  less  than  twice  the  second;  the  fourth  is  twice 
the  first;  the  fifth  is  one-half  the  first;  the  sixth  is  276°.    What 
is  the  number  of  degrees  in  each  angle  of  the  hexagon,  both 
interior  and  exterior? 

8.  Write  a  problem  by  substituting  general  numbers  in- 
stead of  the  specific  numbers  in  Exercise  6  and  solve.    Check 


66  PLANE  GEOMETRY  [n,  §  76 

your  answers  by  substituting  the  special  values  found  in 
Exercise  6,  and  comparing  results  with  answers  to  Exercise  6. 

9.  Repeat  Exercise  8  using  Exercise  7  as  the  basis  of  problem. 

10.  The  point  P  lies  in  Z  CBA.     PQ 
and  PS  are  drawn  perpendicular  to  the 
arms  respectively.     Prove  that  Z  SPQ  is 
the  supplement  of  Z  CBA.  Compare  with 

Theorem  XIX.  B~~  o^c 

11.  What  is  the  value  of  each  angle  of  a  regular  triangle? 
What  is  another  name  for  a  regular  triangle?     What  is  the 
value  of  each  angle  of  a  regular  quadrilateral?    What  is 
another  name  for  a  regular  quadrilateral?      What  is  the 
value  of  each  angle  of  a  regular  pentagon?     Of  a  regular 
hexagon?     Of  a  regular  heptagon?     Of  a  regular  n-gon? 

12.  What  is  the  value  of  each  exterior  angle  of  the  figures 
mentioned  in  Exercise  11? 

13.  What  is  the  ratio  of  the  interior  and  exterior  angle  of 
each  figure  mentioned  above? 

Using  this  ratio  for  the  regular  n-gon  as  a  formula,  find 
the  ratio  of  the  interior  and  exterior  angle  of  a  regular 
polygon  of  12  sides.  Of  20  sides.  Of  50  sides.  Does  the 
ratio  grow  greater  or  less  as  we  increase  the  number  of  sides? 

14.  Draw  the  bisectors  of  the  interior  an- 
gles of  a  regular  hexagon.      Show  that  they 
meet  in  a  point  equidistant  from  the  vertices, 
and  divide  the  hexagon  into  six  equilateral 
triangles. 

15.  Which  of  the  regular  figures  can  be 
arranged  about  a  point  so  as  to  form  designs 
for  wall-paper  or  linoleum?    What  combina- 
tions can  be  so  used?     Make  some  designs 
using  regular  figures.     (In  the  adjacent  de- 
sign three  regular  hexagons  come  together  at  each  vertex. 
The  design  of  Ex.  14  contains  regular  hexagons  and  triangles.) 


II,  §  77]  PARALLELS  AND  TRANSVERSALS  67 

77.  Definitions. 

A  parallelogram  is  a  quadrilateral  whose  opposite  sides 
are  parallel. 

A  rectangle  is  a  parallelogram  which  has  one  angle  a 
right  angle. 

A  square  is  a  rectangle  with  its  adjacent  sides  equal. 

A  rhombus  is  a  parallelogram  whose  sides  are  all  equal, 
but  which  contains  no  right  angle. 

78.  Theorem  XXV.     //  a  convex  quadrilateral  has  its  op- 
posite sides  equal,  the  figure  is  a  parallelogram. 


Given  the  quadrilateral  A  BCD  with 
side  AB  =  side  CD, 
side  DA  =  side  BC. 

To  prove  that  quadrilateral  ABCD  is  a  parallelogram. 

Analysis.  In  order  to  prove  that  quadrilateral  ABCD  is  a 
parallelogram  we  must  prove  that  A  B  is  parallel  to  CD, 
and  that  BC  is  parallel  to  DA.  To  prove  that  these  lines 
are  parallel  we  cross  them  by  a  transversal  AC.  Now  we 
can  prove  AB  \\  CD  if  we  can  prove  that  the  alternate 
interior  angles  BAC  and  DC  A  are  equal,  and  we  can  prove 
that  BC  ||  AD  if  we  can  prove  that  the  alternate  angles  ACB 
and  CAD  are  equal.  We  can  prove  these  angles  equal  if  we 
can  prove  A  ABC  and  A  CD  congruent.  We  can  prove  these 
triangles  congruent  if  we  can  prove  that  three  sides  of  the  one 
are  equal  respectively  to  three  sides  of  the  other. 

Proof.     Let  the  student  give  this  proof  complete. 


68  PLANE  GEOMETRY  [II,  §  79 

79.  Theorem  XXVI.     //  a  convex  quadrilateral  has  two  of 
its  sides  equal  and  parallel,  the  figure  is  a  parallelogram. 

Give  drawing,  analysis,  and  proof  as  in  the  above  Theorem. 

Suggestion.  In  the  quadrilateral  A  BCD  let  A  B  and  CD 
be  parallel  and  equal.  Will  it  then  be  sufficient  to  prove 
BC  =  DA  ?  Can  you  do  this  by  congruent  triangles? 

80.  Theorem  XXVII.     A  diagonal  divides  a  parallelogram 
into  two  congruent  triangles. 


Given  the  parallelogram  A  BCD,  with  the  diagonal  AC, 
forming  triangles  ABC  and  A  CD. 

To  prove  triangles  ABC  and  ACD  congruent. 

Analysis.  We  can  prove  that  A  A  BC  ^  A  ACD  if  we  can 
prove  that  Z  BAG  =  Z  ACD,  and  that  Z  ACB  =  Z  CAD, 
since  line  AC  is  common  to  both.  We  can  show  that  Z  B  AC 
=  Z  ACD  if  we  can  show  that  they  are  alternate  interior 
angles  of  the  parallel  lines  AB  and  CD.  We  can  show  that 
A  ACB  and  CAD  are  equal  if  we  can  show  that  they  are  the 
alternate  interior  angles  of  the  parallel  lines  BC  and  DA. 

Let  the  student  make  this  proof  complete. 

Corollary  1.     The  opposite  sides  of  a  parallelogram  are  equal. 

Corollary  2.     The  opposite  angles  of  a  parallelogram  are  equal. 

Corollary  3.  The  sum  of  two  consecutive  angles  of  a  parallel- 
ogram is  equal  to  a  straight  angle. 

Corollary  4.     Each  angle  of  a  rectangle  is  a  right  angle. 


II,  §  81]  PARALLELS  AND  TRANSVERSALS  69 

81.  Theorem  XXVIII.     The  diagonals  of  a  parallelogram 
bisect  each  other. 

Let  the  student  give  figure,  analysis  and  proof. 

82.  Exercises. 

1.  If  two  lines  bisect  each  other,  the  figure  formed  by  join- 
ing their  ends  is  a  parallelogram. 

2.  The  diagonals  of  a  rectangle  are  equal. 

3.  The  diagonals  of  a  square  are  perpendicular  to  each 
other. 

4.  The  diagonals  of  a  rhombus  are  perpendicular  to  each 
other. 

5.  If  the  four  interior  angles  formed  by  two  parallel  lines 
cut  by  a  transversal  are  bisected,  the  four  bisectors,  extended 
to  intersect,  form  a  parallelogram. 

6.  Prove   that   the   parallelogram   of    Exercise   5   is   a 
rectangle. 

7.  The  bisectors  of  the  angles  of  a  parallelogram  extended  tc 
intersect  form  a  rectangle. 

8.  On  the  diagonal  AC  of  the  parallelogram  A  BCD  equal 
distances  AP  and  CQ  are  laid  off  from  the  vertices  A  and  C. 
The  lines  BQ  and  DP  are  drawn.     Prove  that  the  figure 
PBQD  is  a  parallelogram. 

9.  One  of  the  angles  of  a  parallelogram  is  ?  TT  radians.    What 
is  the  size  of  each  of  the  others?     If  one  is  (™  —  -p  TT  radians, 
what  is  the  size  of  each  of  the  others? 

10.  How  many  radians  would  have  to  be  added  to  the  angle 
given  in  the  first  question  of  Exercise  9,  in  order  that  the 
parallelogram  be  a  rectangle?    How  many  in  the  second 
question? 

11.  One  of  the  angles  of  a  parallelogram  is  m  degrees  more 
than  j  part  of  one  of  the  others.    Find  the  number  of  degrees 
in  each  angle  of  the  parallelogram.     How  many  degrees  if 
m  =  30°,  r  =  3,  s  =  4. 


70  PLANE  GEOMETRY  m,  §  82 

12.  The  longer  side  of  a  rectangle  is  k  cm.  less  than  5  part 
of  the  shorter.     If  h  cm.  be  subtracted  from  the  longer  and 
added  to  the  shorter,  the  figure  will  be  a  square.     Find  the 
length  of  each  side  of  the  rectangle. 

13.  Make  a  square  by  hinging  together  four  strips  of  paper. 
Swing  it  over  until  one  angle  is  f  the  size  of  the  angle  of  the 
square.     What  is  the  size  of  each  angle  of  the  new  figure? 
Give  answer  in  radians  and  also  in  degrees.     What  is  the  new 
figure  called?     What  truths  hold  for  the  new  figure  that  did 
not  hold  for  the  square.     State  truths  that  hold  for  both 
figures. 

14.  If  a  line  is  drawn  through  the  point  of  intersection 
of  the  diagonals  of  a  parallelogram,  parallel  to  one  side,  it 
will  bisect  two  sides  of  the  parallelogram. 

Definitions. 

A  trapezoid  is  a  quadrilateral  which  has  two  parallel  sides, 
the  other  two  sides  being  not  parallel. 

The  parallel  sides  are  frequently  spoken  of  as  the  bases  of 
the  trapezoid. 

If  the  non-parallel  sides  are  equal,  the  trapezoid  is  said  to 
be  isosceles. 

15.  The  angles  at  the  same  base  of  an  isosceles  trapezoid 
are  equal. 

Suggestion.  If  PQRS  be  the  trapezoid,  draw  SM  \\  RQ,  and 
show  that  A  QPS  and  PQR  are  both 
equal  to  Z.  SMP.  This  proves  that 
the  angles  at  the  lower  base  are  equal. 
Now  show  that  the  angles  at  the  up- 
per base  are  equal. 

16.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

17.  Construct  an  isosceles  trapezoid  whose  bases  are  3 
inches  and  4  inches  respectively  and  whose  perimeter  is.  17 
inches. 


n,§83]  PARALLELS  AND  TRANSVERSALS  71 

MORE  THAN  TWO  PARALLEL  LINES  CUT  BY 
TRANSVERSAL 

83.  Theorem  XXIX.     //  a  system  of  parallels  cuts  equal 
parts  on  any  transversal,  it  cuts  equal  parts  on  every  transversal. 


M\ 

Given  the  parallel  lines  I/i,  L2,  I/3,  cutting  the  transversal 
T  in  the  points  Q,  P,  and  M ,  making  segment  m  equal  to 
segment  n;  also  cutting  the  transversal  TI  at  the  points  #, 
A,  and  C,  forming  the  segments  r  and  s. 

We  are  to  prove  that  segment  r  equals  segment  s. 

Analysis.  To  prove  that  segments  r  and  s  are  equal,  we 
may  use  triangles  containing  r  and  s.  In  order  that  such 
triangles  may  be  proved  congruent  they  must  also  contain 
segments  m  and  n  or  the  equals  of  m  and  n.  In  order  to 
accomplish  the  latter  we  draw  the  lines  A  K  and  BH  through 
the  points  A  and  B  respectively  parallel  to  transversal  T, 
forming  the  triangles  CKA  and  AHB.  These  triangles  can 
be  proved  congruent  by  proving  that  two  angles  and  a  side 
of  one  are  equal  respectively  to  two  angles  and  the  corre- 
sponding side  of  the  other. 

Proof. 

What  kind  of  a  figure  is  PHBQ?  Why? 

Then  how  does  line  BH  compare  with  segment  n?  Why? 

What  kind  of  a  figure  is  MKAP1  Why? 

How  does  line  A  K  compare  with  segment  ra?  Why? 


72  PLANE  GEOMETRY  [II,  §  84 

How  do  the  segments  m  and  n  compare?  Why? 

What  can  you  now  say  of  lines  BE  and  A  K1  Why? 

How  do  angles  ABH  and  CA  K  compare?  Why? 

How  do  angles  HAB  and  KCA  compare?  Why? 

What  now  can  you  say  of  A  AHB  and  C KA?  Why? 

What  can  you  state  about  the  line  segments  r  and  s? 

State  the  proposition  proved. 

84.  Theorem  XXX.     A  line  drawn  through  the  mid-point  of 
one  side  of  a  triangle  parallel  to  a  second  side  will  bisect  the 
third  side. 

Suggestion.  Through  the  vertex 
opposite  the  second  side  of  the  tri- 
angle draw  a  line  parallel  to  the  given 
line,  and  apply  Theorem  XXIX. 

85.  Theorem  XXXI.     A  line  drawn  through  the  mid-points 
of  two  sides  of  a  triangle  is  parallel  to  the  third  side. 

Suggestion.  Through  one  of  the  mid-points  draw  a  line 
parallel  to  the  third  side.  According  to  Theorem  XXX 
this  new  line  bisects  the  second  side.  This  new  line  must 
coincide  with  the  line  which  joined  the  mid-points  of  the 
two  sides.  Why?  So  that  both  are  parallel  to  the  third 
side.  Give  this  proof  in  full. 

86.  Theorem  XXXII.    The  line  joining  the  mid-points  of  two 
sides  of  a  triangle  is  equal  to  one-half  of 

the  third  side. 

Suggestion.  D  and  E  are  mid-points 
of  AC  and  BC.  To  prove  DE  =  \  AB, 
take  Fj  the  mid-point  of  A  B  and  draw 
DF.  From  Theorem  XXXI  what  can 
you  say  about  lines  DE  and  A  5? 
About  lines  FD  and  BC?  What  kind  of  a  figure  is  BEDF? 
What  do  you  conclude  about  the  lengths  of  DE  and  ABt 


II,  §  87]  PARALLELS  AND  TRANSVERSALS  73 

87.  Problem  XII.     To  divide  a  given  line-segment  into  n 
equal  parts. 


Given  the  line  segment  AB. 

To  divide  the  line-segment  AB  into  n  equal  parts. 

We  shall  let  n  =5.     The  method  is  the  same  for  any  other 
number. 

Analysis.     Suppose  the  work  done  as  in  the  figure  below. 


AX  is  a  line  of  indefinite  length  making  any  acute  angle 
with  the  given  line  AB.  AM  is  a  segment  whose  length  is 
chosen  at  will.  This  length  is  laid  off  five  times  on  the  line 
AX,  and  the  final  point  is  marked  D.  BD  is  drawn.  The 
lines  through  the  other  points  on  AX  are  drawn  parallel  to 
BD.  These  lines  divide  A B  into  five  equal  parts.  Why? 

Construction.  At  one  end  of  the  line-segment  A  B  draw 
line  AX  making  any  acute  angle  with  A  B.  Starting  at  point 
A  lay  off  on  AX  any  convenient  length  as  AM  as  many  times 
as  the  number  of  divisions  you  wish  A  B  to  contain.  Call  the 
last  point  D.  Draw  BD.  Through  each  point  on  AX  con- 
struct an  angle  equal  to  Z  BDA  making  use  of  Problem  III. 
Extend  the  arms  of  these  angles  to  cut  segment  A  B.  Segment 
A  B  will  be  divided  into  the  desired  number  of  equal  parts. 


Proof.     Let  the  student  give  the  proof,  using  Theorem 
XXIX. 


74  PLANE  GEOMETRY  Hi,  §  88 

88.  Theorem  XXXIII.  //  two  triangles  have  two  sides  of 
the  one  respectively  equal  to  two  sides  of  the  other,  but  the  in- 
cluded angle  of  the  one  greater  than  the  included  angle  of  the 
other  then  the  third  sides  are  unequal,  and  the  greater  angle  has 
the  greater  side  opposite  it. 


Given  the  triangles  ABC  and  AiBiCi  with  side  BC  equal 
to  side  Bid  and  side  CA  equal  to  side  CiAi,  but  angle  ACB 
greater  than  the  angle  AiCiBi. 

To  prove  that  side  AB  is  greater  than  side  AiBi. 

Analysis.  To  prove  this  we  shall  place  the  two  triangles 
together,  placing  A  A^Bid  on  A  ABC,  so  that  Aid  falls  on 
AC.  Then  d#i  will  fall  between  AC  and  CB.  Why?  If 
now  we  can  so  draw  in  a  line  that  we  shall  have  congruent  tri- 
angles, and  at  the  same  time  get  a  connection  between  AB 
and  ABi,  we  can  compare  the  lengths  of  A  B  and  ABi.  To 
do  this  we  bisect  Z  B£B  by  the  line  CD  and  join  BiD. 


We  can  now  show  that  A  B  is  greater  than  ABi,  if  we  can 
show  that  BiD  is  equal  to  DB.  We  can  show  that  BD  is 
equal  to  DB,  if  we  can  show  that  A  B^DC  is  congruent 
to  A  DBC.  We  can  show  that  these  two  triangles  are 
congruent. 


II,  §  89]  PARALLELS  AND  TRANSVERSALS  75 

Proof.                       A  BCD  sa  A  B^DC.  Prove  this. 

DB  =  BtD.  Why? 

But  A  D  +  DBi  >  A  B!  ;  Why? 

AB  >  ABi.  Why? 
State  the  proposition  proved. 

Definition.  When  two  theorems  are  related  to  one  another 
in  such  a  way  that  the  thing  that  is  to  be  proved  in  the  one 
becomes  the  thing  that  is  assumed  to  be  true  in  the  other, 
such  theorems  are  called  converse  theorems.  Either  one  is 
the  converse  of  the  other. 

An  example  of  converse  theorems: 

//  two  sides  of  a  triangle  are  equal  the  angles  opposite  these 
sides  are  equal. 

If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  these 
angles  are  equal. 

State  other  propositions  and  their  converses  that  you  have 
proved. 

The  following  theorem  is  the  converse  of  Theorem 
XXXIII. 

89.  Theorem  XXXIV.  If  two  triangles  have  two  sides  of  the 
one  equal,  respectively,  to  the  two  sides  of  the  other,  but  the  third 
side  of  the  one  greater  than  the  third  side  of  the  other,  then  the  in- 
cluded angles  are  unequal,  the  greater  side  having  the  greater 
angle  opposite  it. 


A  A  B 

Given  the  two  triangles  ABC  and  AiBiCi  with  the  side  BC 
equal  to  side  BiCi  and  the  side  CA  equal  to  side  CiAi,  but  the 
side  AB  longer  than  AiBi. 

To  prove  that  angle  ACB  is  greater  than  angle  AiCiBi. 


76  PLANE  GEOMETRY  [II,  §  89 

Analysis.  To  prove  this,  we  place  the  two  triangles  to- 
gether, placing  AI  on  A,  AiBi  taking  the  direction  of  AB,  the 
two  triangles  on  opposite  sides  of  AB.  Draw  the  line  CCi 
and  extend  CiBi  to  intersect  BC  at  the  point  K. 


Now  since  A  ACCi  and  CCiA  are  equal,  we  can  prove 
Z  ACS  greater  than  Z  BiCiA  if  we  can  prove  Z  CiCB  greater 
than  Z  BiCiC.  We  can  prove  Z  dCB  greater  than  Z  #iCiC, 
if  we  can  prove  that  in  ACCiK  side  C\K  is  greater  than 
side  C  K.  C\K  is  greater  than  C\Bi  which  is  equal  to  CB, 
which  in  turn  is  greater  than  CK. 

Proof.  Z  ACd  =  Z  ACiC.  Why? 

dK>      CiBi.  Why? 

d£i  =       CB.  Why? 

C£  >       CK.  Why? 


/.  ,  in  A  Ci  KC,  Z  dC  K  >  Z  tfCi<7.  Why? 

Z  AC5  >  Z  tfid^.          .      Why? 

State  the  proposition  just  proved. 

Note.    A  proposition  may  be  true  and  its  converse  not 
true. 

For  example: 

//  two  angles  are  right  angles,  their  sum  will  equal  a  straight  angle. 

The  converse, 

//  the  sum  of  two  angles  is  a  straight  angle,  the  angles  are  right 
angles,  is  not  true. 


ii,  §  901  PARALLELS  AND  TRANSVERSALS  77 

90.  Exercises. 

1.  If  the  mid-points  of  the  sides  of  an  equilateral  triangle 
are  joined,  four  congruent  equilateral  triangles  are  formed. 
Prove. 

2.  Start  with  a  given  equilateral  triangle  (rather  large); 
applying  any  of  the  preceding  problems  needed,  divide  it  into 
four  equal  triangles. 

3.  So  construct  that  the  figure  of  Exercise  2  shall  contain 
sixteen  equilateral  triangles. 

4.  If  you  should  divide  each  of  the  equilateral  triangles  of 
Exercise  3  into  four  equilateral  triangles,  how  many  triangles 
would  you  have?     By  another  such  division  how  many  tri- 
angles would  you  have?     Write  down  in  a  row  the  numbers 
of  triangles  in  Exercises  2,  3,  4.     Do  you  find  any  common 
relation  between  these  numbers?    What  is  the  relation?   The 
figure  which  you  have  made  is  frequently  used  as  a  design  for 
linoleum  and  tiling.     See  Exercise  26,  page  52. 

5.  Construct    six    equilateral    triangles    about    a    point. 
Divide  each  into  four  equilateral  triangles.     How  many  tri- 
angles are  there  about  each  point  of  intersection  of  the  lines? 

6.  Make  other  designs  by  bringing  to- 
gether equilateral  triangles  of  various  sizes, 
and  joining  mid-points  of  sides. 

7.  Examine  tiling  floors,  linoleums,  laces 
and  like  forms  to  see  if  the  equilateral  tri- 
angle is  brought  often  into  use. 

8.  If  a  man  sets  five  plants  at  equal  distances  apart  and 
in  a  line  with  the  mid-points  of  two  of  the  border  lines  of  a 
flower  bed  in  the  shape  of  an  equilateral  triangle,  how  many 
plants  can  he  place  the  same  distance  apart  around  the 
border? 

9.  The  mid-point  of  the  hypotenuse  of  a  right-angled  tri- 
angle is  equidistant  from  the  three  vertices.     Prove.     Could 
the  mid-point  of  any  side  of  an  isosceles  triangle  be  equidis- 
tant from  the  three  vertices? 


78 


PLANE  GEOMETRY 


[II,  §  90 


10.  Take  a  card-board  box  and  with  needle  and  thread 
stretch  lines  from  corner  to  corner,  also  from  mid-point  of 
side  to  mid-point  of  side,  as  shown  in  the  figure. 


Assuming  that  the  lines  which  are  parallel  to  the  same 
straight  line  are  parallel  to  each  other  no  matter  whether 
they  are  all  in  the  same  plane  or  not,  prove  the  following 
statements: 

The  diagonals  of  each  face  bisect  each  other; 

The  diagonals  of  the  figure  bisect  each  other; 

The  lines  joining  the  mid-points  of  the  faces  bisect  each 
other,  and  are  parallel  to  the  edges. 

11.  Take  a  card-board  box  and  take  the  bottom  out,  and 
shear  it  to  the  side.     Treat  as  in  Exercise  10.     State  the  facts 
that  are  the  same  as  before,  and  the  variations  that  have  been 
made. 

12.  If  the  sides  of  a  quadrilateral  are  bisected,  the  lines  join- 
ing the  points  of  bisection  in  order  form  a  parallelogram. 

Hint.     Draw  in  the  diagonals  of  the  quadrilateral. 
Note.     When  a  quadrilateral  is  called  for,  be  careful  not 
to  draw  a  special  figure  such  as  a  parallelogram  or  square. 

13.  The  lines  which  join  the  mid-points  of  the  opposite 
sides  of  a  quadrilateral  bisect  each  other. 

14.  Two  straight  lines  are  cut  by  three  parallel  lines.    Call 
the  segments  which  are  cut  on  one  of  the  transversals  a  and  6 
respectively  and  the  segments  cut  on  the  other  c  and  d  re- 
spectively.    If  segment  a  contained  6  units  less,  it  would  be 


II,  §  91]  PARALLELS  AND  TRANSVERSALS  79 

\  as  long  as  segment  c.  If  8  units  were  added  to  twice  seg- 
ment 6,  it  would  be  just  twice  segment  d.  If  segment  q  equals 
segment  6,  find  the  lengths  of  segments  a,  6,  c,  and  d. 

15.  If,  in  Exercise  14,  segment  a  contained  m  units  less  it 
would  be  |  part  of  segment  c.  If  r  units  were  added  to  n 
times  segment  6,  it  would  be  r  times  segment  d.  If  segment 
a  equals  segment  6,  find  the  lengths  of  segments  a,  6,  c,  and  d. 

91.  Summary  of  Chapter  II. 

Part  I.     Angles.     Measurement — Degree  and  Radian. 

Postulate  V.     All  straight  angles  are  equal. 

Corollary.     All  right  angles  are  equal. 

Complementary,  Supplementary,  Conjugate  Angles. 

Theorem  I.     If  two  angles  are  equal,  their  complements  are  equal. 

Theorem  II.     If  two  angles  are  equal,  their  supplements  are  equal. 

Theorem  III.     If  two  angles  are  equal,  their  conjugates  are  equal. 

Vertical  Angles. 

Theorem  IV.     If  two  straight  lines  intersect,  the  vertical  angles  are 
equal. 
Part  II.     Triangles — Congruence  and  Construction. 

Congruent  Triangles. 

Theorem  V.  If  two  triangles  have  two  sides  and  the  included  angle 
of  one  respectively  equal  to  the  sides  and  the  included  angle  of  the  other, 
the  triangles  are  congruent. 

Theorem  VI.  If  two  triangles  have  two  angles  and  the  included  side 
of  the  one  respectively  equal  to  the  two  angles  and  the  included  side  of 
the  other,  the  triangles  are  congruent. 

Isosceles  Triangles.     Equilateral  Triangles.     Definition. 

Theorem  VII.  In  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 

Corollary.     If  a  triangle  is  equilateral  it  is  also  equiangular. 

Perpendicular  lines.     Definition. 

Corollary.  At  a  point  in  a  line  only  one  perpendicular  can  be  erected 
to  that  line. 

Theorem  VIII.  Every  point  in  the  perpendicular  bisector  of  a 
straight  line  is  equidistant  from  the  ends  of  the  line. 

Theorem  IX.  If  two  triangles  have  three  sides  of  the  one  equal  re- 
spectively to  the  three  sides  of  the  other,  the  triangles  are  congruent. 

Constructions. 

Circumference.     Assumptions  with  regard  to  circumferences. 


80  PLANE  GEOMETRY  III,  §  91 

Problem  I.     To  construct  a  triangle  when  three  sides  are  given. 

Problem  II.     To  construct  a  triangle  congruent  to  a  given  triangle. 

This  is  another  form  of  Problem  I. 

Theorem  X.  The  sum  of  two  sides  of  a  triangle  must  be  greater  than 
third  side  and  their  difference  less  than  the  third  side. 

Problem  III.     To  bisect  a  given  angle. 

Corollary  1.  At  a  given  point  on  a  given  line  to  construct  a  per- 
pendicular to  the  given  line. 

Corollary  2.  From  a  given  point  external  to  a  given  line  to  construct 
a  perpendicular  to  the  given  line. 

Problem  IV.  To  construct  the  perpendicular  bisector  of  a  given  line- 
segment. 

Problem  V.  At  a  given  point  on  a  given  line  to  construct  an  angle 
equal  to  a  given  angle. 

Problem  VI.  To  construct  a  triangle  when  two  sides  and  the  in- 
cluded angle  are  given. 

Problem  VII.  To  construct  a  triangle  when  two  angles  and  the  in- 
cluded side  are  given. 

Exterior  angles  of  triangles  and  polygons.     Definition. 

Theorem  XL  The  exterior  angle  of  a  triangle  is  greater  than  either 
interior  angle  not  adjacent  to  it. 

Theorem  XII.  If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  are  unequal,  and  the  greater  side  has  the  greater  angle  op- 
posite it. 

Part  III.     Transversals — Parallel  lines — Sums  of  angles  of  polygons — 

Parallelograms. 

Definitions.  Alternate  interior  angles,  alternate  exterior  angles, 
corresponding  angles. 

Theorem  XIII.  If  two  straight  lines  are  cut  by  a  transversal,  mak- 
ing one  pair  of  alternate  interior  angles  equal,  the  other  pair  of  alternate 
interior  angles  are  equal,  the  pairs  of  alternate  exterior  angles  are  equal, 
the  pairs  of  corresponding  angles  are  equal,  the  sum  of  the  two  interior 
angles  on  the  same  side  of  the  transversal  is  equal  to  a  straight  angle, 
and  the  sum  of  the  exterior  angle  on  the  same  side  of  the  transversal  is 
equal  to  a  straight  angle. 

Corollary.     As  above  starting  with  pairs  of  exterior  angles  equal. 

Definition.     Parallel  lines. 

Postulate  of  Parallels.  Two  intersecting  lines  cannot  be  parallel  to 
the  same  line. 

Theorem  XIV.  If  two  straight  lines  are  cut  by  a  transversal  making 
a  pair  of  alternate  interior  angles  equal  the  lines  are  parallel. 


II,  §  91]  PARALLELS  AND  TRANSVERSALS  81 

Problem  VIII.  Through  a  given  point  to  construct  a  line  parallel 
to  a  given  line.  Involved  in  Theorem  XIII,  XIV. 

Converse  Propositions.     Illustrations  and  definition. 

Theorem  XV.  If  two  parallel  lines  are  cut  by  a  transversal  the 
alternate  interior  angles  are  equal,  etc. 

Theorem  XVI.  A  line  perpendicular  to  one  of  two  parallel  lines  is 
perpendicular  to  the  other. 

Theorem  XVII.  A  line  parallel  to  one  of  two  parallel  lines  is  parallel 
to  the  other. 

Theorem  XVIII.  Two  angles  whose  arms  are  parallel  each  to  each 
are  either  equal  or  supplemental. 

Theorem  XIX.  Two  angles  whose  arms  are  perpendicular  each  to 
each  are  either  equal  or  supplemental. 

Theorem  XX.     The  sum  of  the  angles  of  a  triangle  is  a  straight  angle. 

Corollary  1 .  The  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the 
interior  angles  not  adjacent  to  It. 

Corollary  2.  A  triangle  can  have  but  one  right  angle  or  one  obtuse 
angle. 

Corollary  3.  If  two  triangles  have  two  angles  of  the  one  equal  re- 
spectively to  two  angles  of  the  other,  the  third  angles  are  equal. 

Corollary  4-  If  two  triangles  have  two  angles  and  a  side  of  one  equal 
respectively  to  two  angles  and  the  corresponding  side  of  the  other,  the 
triangles  are  congruent. 

Corollary  5.  If  two  right  triangles  have  an  angle  and  a  side  of  one 
equal  respectively  to  an  angle  and  the  corresponding  side  of  the  other 
the  triangles  are  congruent. 

Theorem  XXI.  If  two  angles  of  a  triangle  are  equal  the  sides  op- 
posite those  angles  are  equal. 

Problem  IX.  Given  two  angles  of  a  triangle,  to  construct  the 
third  angle. 

Problem  X.  To  construct  a  triangle,  given  two  angles  and  a  side 
opposite  one  of  them. 

Problem  XL  To  construct  a  triangle,  given  two  sides  and  an  angle 
opposite  one  of  them. 

Theorem  XXII.  If  two  angles  of  a  triangle  are  unequal  the  sides 
opposite  those  angles  are  unequal  and  the  greater  angle  has  the  greater 
side  opposite  it. 

Corollary  1 .  Of  all  lines  drawn  to  a  given  straight  line  from  a  given 
external  point,  the  perpendicular  is  the  shortest. 

Corollary  2.  The  hypotenuse  of  a  right  angled  triangle  is  the  longest 
side  of  the  triangle. 

Corollary  3.    Of  two  oblique  lines  drawn  to  a  given  line  from  a  given 


82  PLANE  GEOMETRY  [II,  §  91 

external  point  that  which  cuts  off  the  greater  distance  from  the  foot 
of  the  perpendicular  is  the  longer. 

Theorem  XXIII.  The  sum  of  the  interior  angles  of  a  polygon  of  n 
sides  is  (n  —  2)  straight  angles. 

Theorem  XXIV.  The  sum  of  the  exterior  angles  of  any  polygon  is  2 
straight  angles. 

Parallelograms. 

Definitions. — Parallelogram,  rectangle,  square,  rhombus,   diagonal. 

Theorem  XXV.  If  a  convex  quadrilateral  has  each  pair  of  its  op- 
posite sides  equal  the  figure  is  a  parallelogram. 

Theorem  XXVI.  If  a  convex  quadrilateral  has  one  pair  of  opposite 
sides  equal  and  parallel,  the  figure  is  a  parallelogram. 

Theorem  XXVII.  A  diagonal  divides  a  parallelogram  into  two 
congruent  triangles. 

Corollary  1.     The  opposite  sides  of  a  parallelogram  are  equal. 

Corollary  2.     The  opposite  angles  of  a  parallelogram  are  equal. 

Corollary  3.  The  sum  of  any  two  consecutive  angles  of  a  parallel- 
ogram is  equal  to  a  straight  angle. 

Theorem  XXVIII.  The  diagonals  of  a  parallelogram  bisect  each 
other. 

More  than  two  parallel  lines  cut  by  transversal. 

Theorem  XXIX.  If  a  system  of  parallels  cuts  equal  parts  on  any 
transversal,  it  does  on  every  transversal. 

Theorem  XXX.  If  a  line  is  drawn  through  the  mid-point  of  one 
side  of  a  triangle,  parallel  to  a  second  side,  it  bisects  the  third  side. 

Theorem  XXXI.  If  a  line  is  drawn  through  the  mid-points  of  two 
sides  of  a  triangle,  it  is  parallel  to  the  third  side. 

Theorem  XXXII.  If  a  line  is  drawn  through  the  mid-points  of  two 
sides  of  a  triangle,  it  is  equal  to  one-half  of  the  third  side. 

Problem  XII.     To  divide  a  line-segment  into  n  equal  parts. 

Theorem  XXXIII.  If  two  triangles  have  two  sides  of  the  one  re- 
spectively equal  to  two  sides  of  the  other,  but  the  included  angles 
unequal,  then  the  third  sides  are  unequal,  the  greater  side  being  opposite 
the  greater  angle. 

Theorem  XXXIV.  If  two  triangles  have  two  sides  of  the  one  re- 
spectively equal  to  two  sides  of  the  other,  but  the  third  sides  unequal, 
then  the  included  angles  are  unequal,  the  greater  angle  being  opposite 
the  greater  side. 

1.  Into  what  three  parts  is  this  chapter  divided  as  to  subject  matter? 

2.  Are  the  subjects  entirely  independent? 

3.  Read  carefully  the  theorems  of  Part  II,  and  state  those-  (if  any) 


II,  §  911  PARALLELS  AND  TRANSVERSALS  83 

which  were  proved  independently  of  Part  I.     State  those   (if  any) 
which  depend  for  their  proof  on  theorems  in  Part  I. 

4.  Go  through  the  theorems  of  Part  III,  and  state  those  that  depend 
on  the  theorems  in  Part  I,  then  those  that  depend  on  the  theorems  in 
Part  II,  then  those  that  depend  on  theorems  from  both  of  these. 

5.  What  two  things  are  involved  in  the  statement  of  a  theorem? 

6.  What  directs  you  in  the  drawing  of  a  figure? 

7.  What  directs  you  as  to  what  you  shall  state  is  given?    As  to 
what  you  shall  state  is  to  be -proved? 

8.  What  is  the  object  of  an  analysis? 

9.  In  general  how  do  the  steps  in  the  proof  correspond  to  the  steps 
in  the  analysis? 

10.  Read  over  the  list  of  exercises  and  pick  out  those  that  seem 
to  you  to  have  real  application,  that  is,  seem  directly  related  to  real 
things  around  you. 

In  the  history  at  the  opening  of  this  book  it  was  stated  that  the 
geometry  which  Euclid  wrote,  300  B.  C.,  was  divided  into  thirteen  books. 
The  theorems  of  this  chapter  are  selected  from  Book  I. 


CHAPTER  III 
EQUALITY  OF  FIGURES 

92.  Equal  Figures.*    If  you  draw  a  geometric  figure  of 
any  shape,  cut  it  up  and  put  the  parts  together  in  a  different 
order,  the  new  figure  will  differ  from  the  old  in  form,  but  not 
in  size  or  area. 

Definition.  Figures  having  the  same  size  are  called  equal 
figures. 

Addition  and  subtraction  of  polygons.  Two  convex  poly- 
gons may  be  added  by  placing  one  without  the  other  with  a 
portion  of  their  perimeters  together  and  then  erasing  that 
portion  of  their  perimeters.  The  new  polygon  will  be  the 
sum.  Two  convex  polygons  may  be  subtracted  by  placing 
one  within  the  other.  The  difference  between  their  areas 
can  readily  be  decided. 

A  polygon,  such  as  a  star-shaped  figure,  can  easily  be 
divided,  if  necessary,  into  several  convex  polygons  by  draw- 
ing in  one  or  more  lines. 

93.  Definitions. 

Either  set  of  parallel  sides  of  a  parallelogram  may  be  re- 
garded as  the  bases,  in  which  case  the  perpendicular  distance 
between  the  two  parallel  lines  is  the  corresponding  altitude. 

Any  side  of  a  triangle  may  be  regarded  as  its  base,  then 
the  perpendicular  from  the  opposite  vertex  to  the  chosen 
side  is  the  corresponding  altitude. 

The  parallel  sides  of  a  trapezoid  are  its  bases,  and  the  per- 
pendicular distance  between  the  two  bases  is  the  altitude. 

*  In  connection  with  this  chapter,  First  Course,  pages  136-144  may  be  reviewed. 

84 


in,  §  94]                              EQUAL  FIGURES  85 

94.  Theorem  I.      Parallelograms  on  the  same  base,  or  on 
equal  bases,  and  between  the  same  parallel  lines  are  equal 

D         D,    c         c,                  C-D,           c,    D             c      ,  c 


V  V  \/\/ 

A  B  A  B 


Given  the  parallelograms  A  BCD  and  ABCiDi  standing 
on  the  same  base  A  B  and  between  the  same  parallel  lines. 

To  prove  the  parallelogram  ABCD  equal  to  the  parallelogram 
ABdD,. 

Note.  The  three  figures  show  the  three  ways  in  which 
the  parallelograms  can  stand  with  reference  to  each  other. 

Analysis.  HJ  ABCD  and  ABCiDi  are  not  the  same  shape, 
but  we  can  show  them  to  be  equal  if  we  can  show  that  for 
every  portion  of  the  one  there  is  an  equal  portion  of  the 
other.  In  the  first  figure  parallelogram  ABCD  is  divided 
into  the  two  portions,  triangle  ADiD  and  the  trapezoid 
ABCDi,  while  parallelogram  ABCiDi  is  divided  into  triangle 
BCiC  and  trapezoid  ABCDi.  Since  the  trapezoid  ABCDi  is 
common  to  both,  we  have  but  to  prove  that  A  ADiD  is  con- 
gruent to  A  BCiC.  A  ADiD  is  congruent  to  A  BCiC  if  we 
can  prove  that  AD  equals  BC,  that  ADi  equals  BCi  and  that 
included  angle  DiAD  equals  included  angle  CiBC. 

Proof.    In  the  AAD^trndBdC 

AD  =  BC.  Why? 

ADi  =  BCi.  Why? 

Z  D,AD  =  Z  CiBC.     (Art.  62)  Why? 

A  AD^D  &  A  BdC.  Why? 

A  A  DiD  +  trapezoid  A  BCDl 

=  A  BdC  +  trapezoid  A  BCDlf          Why? 
H  ABCD  =  a  ABdDi.  Why? 

Give  analysis  and  proof  for  each  of  the  other  figures. 
State  the  proposition  proved. 


86  PLANE  GEOMETRY  [in,  §  95 

Corollary  1.     Parallelograms  having  equal  bases  and  equal 
altitudes  are  equal. 

Corollary  2.     A  parallelogram  equals  a  rectangle  which  has  the 
same  base  and  the  same  altitude. 

95.  Theorem  II.     Triangles  which  stand  on  the  same  base 
and  are  between  the  same  parallel  lines  are  equal. 


Given  the  triangles  ABC  and  ABCi  standing  on  the  same 
base  and  between  the  same  parallel  lines. 

To  prove  triangle  ABC  equal  to  triangle  ABCi. 

Analysis.  Since  we  know  that  parallelograms  are  equal 
if  they  have  the  same  bases  and  are  between  the  same  par- 
allel lines,  we  can  show  that  A  ABC  and  ABCi  are  equal  if 
we  can  show  that  they  are  halves  of  such  parallelograms.  To 
do  this  draw  AP  parallel  to  side  BC,  and  BQ  parallel  to  side 
A  Ci  forming  [U  A  BCP  and  A  BQCi  respectively.  Then  show 
that  A  ABC  is  one-half  of  O  A  BCP,  and  that  A  ABCi  is 
one-half  of  O  A  BQCi. 

Proof.    Let  the  student  give  this  proof. 

Corollary  1.  Triangles  which  have  equal  bases  and  equal 
altitudes  are  equal. 

Corollary  2.  A  triangle  equals  half  a  parallelogram  which 
stands  on  the  same  base  and  lies  between  the  same  parallel  lines. 
This  might  be  stated. 

Corollary  3.  A  triangle  is  equal  to  one-half  of  a  parallelogram 
which  has  the  same  base  and  altitude. 


in,  §  96]  EQUAL  FIGURES  87 

Exercise.  A  parallelogram  is  divided  into  four  triangles 
by  drawing  in  the  diagonals.  Compare  their  areas. 

96.  Theorem  III.  A  trapezoid  is  equal  to  one-half  of  a 
rectangle  whose  base  is  the  sum  of  the  parallel  sides  and  whose 
altitude  is  the  same  as  the  altitude  of  the  trapezoid. 


S  R 


P  Q  p 

Given  the  trapezoid  PQRS. 

To  prove  that  the  trapezoid  PQRS  is  equal  to  one-half  of  a 
rectangle  whose  base  is  equal  to  PQ  plus  RS  and  whose  altitude 
is  the  same  as  that  of  the  trapezoid. 

Analysis.  We  shall  first  prove  that  the  trapezoid  PQRS 
is  equal  to  half  the  parallelogram  whose  bases  are  the  sum 
of  PQ  and  RS,  and  whose  altitude  is  the  same  as  that 
of  the  trapezoid.  In  order  to  have  such  a  parallelogram, 
take  a  trapezoid  identical  to  trapezoid  PQRS,  invert  it 
and  place  it  so  that  side  QR  falls  on  side  RQ  as  shown  in 
the  figure. 

Call  the  new  figure  P&PiS.  Figure  PS^S  will  be  the 
required  parallelogram  provided  that  we  can  prove  that  QSi 
falls  in  a  straight  line  with  PQ  and  that  RPi  falls  in  a  straight 
line  with  SR,  and  that  PiSi  is  equal  and  parallel  to  SP.  We 
can  prove  that  QSi  falls  in  a  straight  line  with  PQ  if  we  can 
prove  that  Z  RQP  +  Z  SiQR  equals  a  straight  angle,  and 
we  can  prove  that  line  RPi  falls  in  a  line  with  SR  if  we  can 
prove  that  Z  SRQ  +  Z  QRPi  equals  a  straight  angle.  Since 
trapezoid  QSiPiR  is  the  trapezoid  inverted,  this  amounts 
to  proving  that  Z  SRQ  +  Z  RQP  is  equal  to  a  straight 
angle. 


88  PLANE  GEOMETRY  [III,  §  97 

Proof.     Z  SRQ  +  Z  RQP  =  a  straight  angle.  Why? 

Z  SiQR  =  Z  SRQ.  Why? 

Z  &QP  +  Z  #QP  =  a  straight  angle.  Why? 

PQ  and  QSi  are  in  a  straight  line.  Why? 

In  like  manner  R PI  is  in  a  straight  line  with  SR. 

PS,  ||  SPi.  Why? 

P&  =  SPL  Why? 

PSiPiS  is  a  parallelogram.  Why? 

Trapezoid    PQRS   is    one-half   of   parallelogram 
PSiPiS.  Why? 

Therefore  trapezoid  PQRS  is  equal  to  one-half  of  a  rec- 
tangle whose  base  is  PQ  +  RS  and  whose  altitude  is  the 
same  as  that  of  the  trapezoid.  Why? 

State  the  proposition  proved. 

97.  From  experiments,  as  in  First  Course,  page  22,  §  16,  it 
can  easily  be  brought  out  that  the  number  of  square  units  in 
the  area  of  a  rectangle  is  equal  to  the  number  of  linear  units 
in  the  base  multiplied  by  the  number  of  linear  units  in  the 
altitude.  Briefly  stated,  the  area  of  a  rectangle  equals  the 
product  of  its  base  by  its  altitude. 

For  the  present  we  shall  assume  this.  A  formal  proof  will 
be  given  later. 

Corollary  1.  The  area  of  a  parallelogram  is  equal  to  the 
product  of  its  base  by  its  altitude.  (It  must  be  remembered  in 
these  statements  that  it  is  numbers  that  are  equal,  namely, 
the  number  of  square  units  of  area  is  equal  to  the  product  of 
the  number  of  linear  units  in  the  base  by  the  number  of  linear 
units  in  the  altitude.) 

Corollary  2.  The  area  of  a  triangle  is  equal  to  one-half  of  the 
product  of  the  base  by  the  altitude. 

Corollary  3.  The  area  of  a  square  is  equal  to  the  square  of  one 
of  its  sides. 


in,  §  98,  EQUAL  FIGURES  89 

Corollary  4.  The  area  of  a  trapezoid  is  equal  to  one-half  of  the 
product  of  the  sum  of  its  bases  by  the  altitude. 

The  above  may  be  expressed  algebraically, 

for  the  rectangle  a  =  bh} 

for  the  parallelogram  a  =  bh, 

for  the  triangle  a  =  \bh, 

for  the  trapezoid  a  =  \(b\  +  b2)h, 

where  a,  b  and  h  equal  the  number  of  units  in  the  area,  base, 
and  altitude  of  the  figures  respectively. 

98.  Exercises.    Algebraic  and  Geometric 

1.  The  length  of  a  rug  is  5  feet  and  the  width  is  3J  feet. 
What  is  the  number  of  square  feet  in  its  area? 

2.  The  design  of  the  rug  spoken  of  in  Exercise  1  is  a  border 
\  foot  wide  about  a  plain  center.     Find  the  area  of  the  center 
and  also  the  border. 

3.  It  takes   103  square  feet  and   128  square  inches  of 
linoleum  to  cover  a  hallway.     The  hall  is  6  feet  8  inches 
wide.     How  long  is  it? 

4.  If  a  rectangle  is  \ob  square  units  in  area,  and  fa  units  in 
width,  how  long  is  it? 

5.  If  a  parallelogram  is  |  +  J-  square  units  in  area,  and 
J-  —  -f  units  in  width,  how  long  is  it? 

Note.  If  the  student  finds  that  he  cannot  handle  his  frac- 
tions readily  he  should  look  up  the  work  in  a  text-book  on 
Algebra,  as  First  Course,  pages  150  to  156  inclusive. 

6.  If  a  triangle  is  f  a  +  f  b  square  units  in  area  and  J  a 
—  -f%  b  units  in  altitude,  what  is  the  number  of  units  in  the 
base? 

7.  A  trapezoid  has  an  area  of  1J  r  —  2f  s  square  units.    One 
of  its  bases  is  If  r  +  T\  s  —  3  and  the  other  base  is  f  r  —  T9f  s 
+  3  units.     Find  the  altitude. 

8.  A  cut  through  a  hill  is  25  feet  wide  at  the  top  and  12  feet 
wide  at  the  bottom.     It  has  a  depth  of  10  feet.     What  is  the 
area  of  a  cross  section? 


90  PLANE  GEOMETRY  [in,  §  99 

9.  Measure  the  width  and  height,  including  the  sloping 
roof,  of  a   bird  house   and  compute   the   area  of  a  cross 
section. 

10.  Measure  the  distance  across  the  top  and  bottom,  and 
the  height  of  a  cork.     Compute  the  area  of  a  section  through 
the  axis  of  the  cork. 

11.  Points  P  and  Q  are  any  points  in  the  sides  AB  and  CD 
respectively  of  O  A  BCD.     The  straight  lines  PD,  PC,  QA, 
and  QB  are  drawn.     Prove  that  A  PCD  equals  A  ABQ. 
Give  both  an  algebraic  and  a  geometric  proof. 

12.  A  line  is  drawn  parallel  to  two  sides  of  a  parallelogram, 
and  cutting  the  other  two  sides.     A  point  P  on  the  line  is 
joined  to  the  vertices  of  the  parallelogram.    Prove  that  the 
sum  of  the  two  triangles  formed  with  the  first  two  sides  of  the 
parallelogram  is  equal  to  one-half  the  area  of  the  parallelo- 
gram.    Give  both  algebraic  and  geometric  proof. 

13.  P  is  any  point  in  the  parallelogram  MNHK.    Lines 
are  drawn  joining  P  to  each  of  the  vertices  of  the  parallelo- 
gram.    Prove  that  the  sum  of  A  MP  K  and  PNH  equals  the 
sum  of  A  MNP  and  PH  K.     Give  both  algebraic  and  geo- 
metric proof.  D c 

14.  Prove  Corollary  4  by 
means  of  the  adjacent  fig- 
ure. 


Note.  In  connection  with  the  following  it  is  advisable  to 
review  First  Course,  pages  127  to  135  inclusive. 

99.  To  shorten  the  naming  of  a  rectangle  it  may  be  named 
diagonally  across.  For  the  same  reason  we  shall  adopt  the 
plan  of  naming  by  the  lengths  of  two  adjacent  sides.  There  is 
a  double  convenience  in  this,  since  the  product  of  these  two 
lengths  gives  the  area. 


Ill,  §  100] 


EQUAL  FIGURES 


91 


100.  Theorem  IV.  The  rectangle  of  two  given  lines  equals  the 
sum  of  the  rectangles  contained  by  one  of  them  and  the  several 
segments  into  which  the  other  is  divided. 


a 
b 

c 

d 

Given  the  rectangle  on  the  lines  A  B  and  AH  with  line  AH 
divided  into  the  segments  6,  c,  d  by  the  points  P  and  Q.  PR 
and  QT  are  lines  drawn  parallel  to  AB.  Line  AB  is  a  and 
line  AH  is  b  +  c  +  d. 

To  prove  that  a(b  +  c  -f  d)  =  ab  +  ac  +  ad. 

Analysis.  Since  rectangle  a  (b  +  c  +  d)  is  made  up  of  n 
APRB,  PQTR  and  QHMT,  we  have  but  to  prove  that  PR 
and  Q77  are  each  equal  to  a,  so  that  n  APRS  =  n  a&, 

PQ7\R  =  n  ac,  and  en  QHMT  =  n  ad. 

Proof.     Let  the  student  give  this  proof  in  full. 

101.  Exercises.  Write  the  dimensions  of  the  rectangles 
which  are  composed  of  the  following  sums: 

1.  mh  +  mk  +  mr.  4.  4  r  +  28. 

2.  p2  +  pq  +  2p.  6.  25  +  30  +  45. 

3.  tz  +  3t  +  ts.  6.  2ras  +  4rs  +  6s. 
Notice  that  we  have  here  simply  the  algebraic  process 

of  factoring,  illustrated  geometrically.  Thus  the  rectangle 
whose  area  is  3  hm  +  9  hn+  6  hq  can  be  regarded  as  a  sum 
of  three  rectangles,  namely  3  hm,  9  hn,  and  6  hq.  These 
can  be  regarded  as  having  a  common  dimension  3  h,  thus : 
3  h  •  m,  3  h  •  3  n,  and  3  h  •  2  g.  Their  sum  then  is  a  rec- 
tangle 3h  •  (m  +  3n  +  2g),  which  is  the  factored  form  of 
the  original  expression. 


92 


PLANE  GEOMETRY 


[III,  §  102 


102.  Theorem  V.  The  square  on  the  sum  of  two  line  segments 
is  equal  to  the  sum  of  the  squares  on  these  segments  plus  two  times 
their  rectangle. 


b          ab               b 

tf 

a               H 
a           as             a 

b 
ab 

a 

b 

Given  the  line  segments  a  and  6,  and  the  square  A  BCD  on 
the  sum  of  these  two  lines. 

To  prove  that  (a  +  &)2  =  a2  +  2ab  +  b2. 

Draw  KE  parallel  to  AD,  and  FG  parallel  to  AB,  cutting 
KE  in  point  H.  Analyze  and  prove. 

103.  Exercises.  Write  the  side  of  the  squares  which  have 
the  following  areas  : 

4.  225  +  30  r  -f  r2. 

5.  625  +  100  p  +  4  p\ 

6.  49  s2  +  42  ps  +  9  p2. 

The  rectangle  of  a  given  line  and  the 


1.  m2  +  4  m  +  4. 

2.  9p2  +  12pg  + 

3.  25  +  40  +  16. 

104.  Theorem  VI. 


difference  between  two  other  given  lines  is  equal  to  the  difference 
between  the  rectangles  of  the  one  line  and  each  of  the  others. 


b-c 


Ill,  §  105] 


EQUAL  FIGURES 


93 


Given  a,  6,  c  the  length  of  three  lines,  and  the  rectangle 

of  a  and  (b  —  c). 

To  prove  that  a(b  —  c)  =  ab  —  ac. 

Let  the  student  give  analysis  and  proof. 

105.  Exercises.     Write  the  dimensions  of  the  rectangle 
which  is  the  difference  between  the  following  rectangles. 


1.  2  a  -  2  c. 

2.  5  r  -  25. 

3.  62  -  be. 


4.  p 


2  _ 


pq. 


5.  2hk-8k*-k. 

6.  3  s2  -  27  rs  +  9  st. 


106.  Theorem  VII.  TTie  square  on  the  difference  between  two 
lines  equals  the  sum  oj  the  squares j)n  the  two  lines  minus  two 
times  their  rectangle.  f 


K                               M 

(o-b)  * 

b 

R 

" 

a-6 

b 

b 

C                           B 

Given  the  lines  a  and  6,  with  RHMK  the  square  on  their 
difference. 

If  line  A  Bis  a  and  line  AC  is  6,  then  the  square  RHMK  is 
(a  -  6)2. 

To  prove  that   (a  -  b)2  =  a2  +  b2  -  2ab. 

Analysis.  To  prove  that  (a  —  b)2  =  a2  +  b2  —  2  ab,  we 
prove  that  ABMN  is  the  square  on  a,  that  n  ABHD  which  is 
subtracted  from  ABMNj^jn  ab,  tliat  ACRD  is  the  square  on 
6  and  must  be  added  before  we  can  subtract  a  ACKN,  which 
is  a  ab.  Thus  we  are  to  show  that  RHMK  is  the  result  of 
subtracting  from  a2  the  rectangle  ab,  then  adding  62,  and 
then  subtracting  rectangle  ab.  This  will  prove  our  statement. 

Proof.    Let  the  student  give  this  proof. 


94 


PLANE  GEOMETRY 


[III,  §  107 


107.  Exercises.     Write  the  side  of  the  square  whose  area  is 
as  follows : 

1.  9-12  +  4.  4.  t*  -  30  ts  +  225  s2. 

2.  16  r2  -  40  rs  +  35  s2.  5.  121  62  -  66  b  +  9. 

3.  81  p2  -  ISp  +  1.  6.  49  -  70  +  25. 

108.  Theorem  VIII.     The  difference  of  the  squares  of  two  lines 
is  equal  to  the  rectangle  of  the  sum  and  difference  of  the  lines. 


D 

F              P 

\ 

a 

b 

R 

b 

a-h 

•t. 

A 

C                             B 

Given  the  segments  A  B  =  a,  AC  =  b,  SD  their  difference 
and  SM  their  sum.  Then  rectangle  SMPD  is  the  rectangle 

of  (a  +  6)  (a  -  6). 

To  prove  that  a2  -  b2  =  (a  -f  &)  (a  -  &). 

Analysis.  Since  ABFD  is  a2  and  ACRS  is  62,  then  the  ir- 
regular figure  SRCBFD  is  the  difference  between  a2  and  62. 
We  must  prove  that  n  SMPD  is  equal  to  irregular  figure 
SRCBFD. 

Proof.     Let  the  student  give  this  proof  in  full. 

109.  Exercises.  Write  the  dimensions  of  the  rectangles 
equal  to  the  following  differences  of  squares. 

1.  16-9.  4.  121d2-49c2. 

2.  25  a2  -  36.  5.  4  h2  -  36. 

3.  144m2  -  n2.  6.  36  -  1. 

7.  By  the  same  plan  of  proof  as  given  for  Theorem  V 
prove  that 

a2  +  am  -f  an  +  mn  =  (a  +  m)  (a  +  n). 
Check  by  substituting  numbers  for  a,  m,  and  n.  , 


in,  §  109]  EQUAL  FIGURES  95 

8.  Write  the  dimensions  of  the  rectangle  composed  of 
s2  +  5  s  +  6. 

9.  By  the  same  plan  of  proof  as  given  in  Theorem  VII  prove 
that 

a2  —  am  —  an  +  mn  =  (a  —  m)  (a  —  n). 

10.  By  the  same  plan  prove  that 

a2  +  am  —  an  —  mn  =  (a  —  n)  (a  +  m) . 

11.  By  drawing,  find  the  side  of  the  rectangle  whose  area  is 
expressed  by 

a2  +  2  ab  +  62  -  ra2  -f  2  mn  -  n2. 
Suggestion:     Note  that  the  given  expression  is  equal  to 

(a2  +  2  ab  +  62)  -  (m2  -  2  mn  +  n2), 
which  is  equal  to 

(a  +  6)2  -  (m  -  ri)\ 
To  this  apply  Theorem  VIII,  page  94. 
Check  by  substituting  numbers  for  a,  6,  m,  n. 

12.  Write  the  dimensions  of  the  following  rectangles: 

(a)  r2  -  7  rs  +  10  s2.  (e)    3  a2  +  10  a  +  3. 

(b)  /b2-2/b-8.  (f)    5r2-7r  +  2. 
(e)   6  m2  -  5  m  -  1.  (g)   4  m2  -  5  m  -  9. 
(d)   1  -  p*  -  2  pr  -  r2.          (h)  7  p2  +  11  ps  -  6  s2. 

13.  It  required  90  square  blocks  of   tiling   to  make  a 
floor  for  a  hallway.     It  was  necessary  to  use  3  more  than 
twice  as   many  in  the  length  as  in  the  width.     If  each 
block  was  a  foot  square,  how  long  and  how  wide  was  the  hall? 

14.  Plants  are  arranged  in  the  form  of  a  hollow 
rectangle  at  uniform  distance  apart.     Along  the 
outer  row  there  are  18  one  way  and  11  the  other. 
How  many  rows  will  it  take  for  138  plants? 

15.  In  a  strip  of  wall-paper  (design  of  flowers  on  a  plain 
ground)  there  is  one  more  than  3  times  as  many  in  the  length 
as  in  the  width.     There  were  14  flowers  on  the  strip.     How 
long  is  the  strip  if  the  flowers  are  8  inches  apart? 


96 


PLANE  GEOMETRY 


[in,  §  no 


110.  Theorem  IX.  The  square  on  the  hypotenuse  of  a 
right  triangle  is  equal  to  the  sum  of  the  squares  on  the  other  two 
sides. 


Given  the  right  triangle  ABC  with  B  the  right  angle,  and 
$,  Si,  $2  the  squares  on  the  hypotenuse  and  sides  respect- 
ively. 

To  prove  that  S  =  Si  +  S2. 

Analysis.  To  make  this  proof  we  divide  the  square  S  into 
two  parts  by  drawing  a  line  BD  through  B  parallel  to  AF,  the 
side  of  square  S.  If  we  can  now  prove  that  a  A  EDF  is  equal 
to  square  Si,  and  that  n  ECLD  is  equal  to  square  S2,  we  shall 
have  proved  our  theorem.  Consider  n  FA  ED  and  square  Si 
first.  We  can  prove  these  equal  if  we  make  two  triangles,  one 
with  a  base  and  altitude  equal  to  those  of  n  FAED,  the  other 
with  base  and  altitude  equal  to  those  of  square  Si,  and  then 
prove  these  triangles  congruent.  In  order  to  have  these  tri- 
angles draw  the  lines  BF  and  CH. 


Proof.    In  &ABFandAHC, 
AF  =  AC. 
AB  =  AH. 
Z  BAF  =  Z  HAC. 


Why? 
Why? 
Why? 


Ill,  §  111] 


EQUAL  FIGURES 


97 


Why? 
Why? 

Why? 

Why? 


(Notice  of  what  two  parts  each  angle  is  the  sum.) 
A  FAB  &  A  ARC. 
[=)FAED  =2  AFAB. 
Square  Si  =  2  A  AHC. 
(Explain  why  CB  and  HA  are  parallel.) 

aFAED  =  square  Si. 
Let  the  student  give  a  complete  proof  that 
n  ECLD  =  square  S2. 
tnFAED  +  cn  ECLD  =  square  Si  +  square  &. 

Why? 

square  5  =  square  Si  +  square  $2. 
State  the  proposition  proved. 

This  theorem  is  called  the  Pythagorean  Theorem  from 
Pythagoras  (580-500  B.C.)  who  first  proved  it.   Many  proofs 
have  been  given  since  his  time.     The  one 
given  here  is  attributed  to  Euclid  (about 
300  B.  C.)«     Pythagoras  used  the  adja- 
cent figure  for  his  proof.     It  is  supposed 
to  have  been  suggested  to  him  by  the  pat- 
tern in  a  tile  floor.     This  theorem  was 
known  many  centuries  before  the  time  of 
Pythagoras,  but  it  had  not  been  proved. 

Is  this  figure  general?     Give  a  proof  using  this  figure. 

111.  Prove  the  Pythagorean  Theorem  using  the  following 
figures.  Here  A  ABC  is  shown 
with  the  squares  on  its  three 
sides.  In  the  second  figure 
squares  I  and  II  are  shown  side 
by  side,  with  dotted  lines  to  a 
point  R  so  placed  that  AR 
equals  the  side  BC  of  A  ABC. 
The  third  figure  shows  square 
III, with  AS  and  SF  drawn  in  so 
as  to  make  A  AFS  e*  A  ABC 


98  PLANE  GEOMETRY 

of  the  first  figure.     Then  G T  is  drawn  _L  FS. 


[in,  §  112 


E                       D 

\ 

H                  B 

X 

\  I 

,»' 

m    >T 

\ 

x'n 

Draw  the  second  figure,  taking  A  ABC  in  the  first  figure 
with  sides  AC  =  2  inches,  BC  =  1.5  inches,  A  B  =  2.5  inches. 
Cut  your  drawing  into  three  parts  by  cuts  along  the  dotted 
lines. 

Draw  the  third  figure,  using  same  dimensions  as  above. 

Show  that  the  three  parts  into  which  you  cut  the  second 
figure  will  exactly  cover  the  third  figure. 

square  I  +  square  II  =  square  III. 

See  also  First  Course,  pages  170-171. 

112.  Problem  I.  To  construct  a  right  triangle  when  the  two 
sides  are  given. 

Y 


Given  the  segments  a  and  b. 

To  construct  a  right  triangle  with  a  and  b  as  sides. 

Construction.  Draw  AB  =  a.  At  B,  draw  BY  _L  AB. 
On  BY  lay  off  BC  =  b.  Draw  AC.  Then  A  ABC  is  the  re- 
quired triangle. 


Ill,  §  113] 


EQUAL  FIGURES 


Proof.     Let  the  student  give  the  proof,  first  making  an 
accurate  construction. 

113.  Problem  II.     To  construct  a  right  triangle  when  one 
side  and  the  hypotenuse  are  given. 

Y 


Given  the  side  equal  to  a  and  the  hypotenuse  equal  to  6. 

To  construct  a  right  triangle. 

Construction.  Draw  AB  =  a,  and  at  B  draw  BY  _L  A B. 
With  A  as  center  and  compass  opened  to  the  distance  b,  strike 
an  arc  cutting  BY  at  C.  Then  A  ABC  is  the  required  tri- 
angle. 

Proof.     Student  supply  proof. 

Discussion.    When  will  such  a  construction  be  impossible? 

114.  Problem  III.  To  construct  a  square  which  will  be  the 
sum  of  two  given  squares. 


Given  the  squares  Si  and 

To  construct  a  square  equal  to  the  sum  of  Si  and  S2. 


100  PLANE  GEOMETRY  [in,  §  us 

Let  the  student  give  a  description  of  this  construction,  with 
reasons. 

115.  Problem  IV.     To  construct  the  square  root  of  a  number. 

We  shall  suppose  the  number  to  be  a  positive  integer.  If 
it  is  a  perfect  square,  the  construction  can  be  made  at  once. 

When  the  number  is  not  a  perfect  square,  we  first  express 
it  as  the  sum  of  two  or  more  squares,  and  then  use  the 
Pythagorean  Theorem. 

Example  1.     Construct  a  line-segment  of  length  -y/13. 

13  =  22  +  32.  c 

Take  any  unit  of  length. 
Draw  AB  =  2  units. 
DrawBC  =  3  units,  and  _L  AB. 
Then  AC  =  \/13  units.     Why? 
Example  2.     Construct  a  line-segment  of  length 


62  =  22  +  32  +  72. 

Take  any  unit  of  length. 

Draw  AB  =  2  units. 

Draw  BC  =  3  units,  and  J_  AB. 

Then  AC  =  \/13  units. 

Draw  CD  =  7  units,  and  J_  AC. 

Then  AD  =  \/62  units. 

Example  3.     Construct  a  line-segment  of  length  \/24. 

(a)  Let  24  =  42  +  22  +  22,  and  construct  as  in  Example  2. 

(6)  Let  \/24  =  \/4  X  6  =  2  \/&    Construct  \/6^  and  double 

the  result, 
(c)  Let  24  =  52  —  I2. 

Construct  a  right  triangle  with  a  side  1  unit  long,  and  hypot- 
enuse 5  units  long.  See  Art.  113.  The  other  side  of  this 
triangle  will  equal  \/24  units.  Explain. 

116.  Exercises.    Algebraic  and  Geometric. 

1.  Find  the  side  of  a  square  which  equals  the  sum  of  two 
given  squares  whose  sides  are  4  and  7  respectively;  13  and  8. 

2.  Construct  a  square  which  is  twice  the  size  of  a  given 
square. 


in,  §  116]  EQUAL  FIGURES  101 

3.  What  is  the  ratio  between  the  side  of  the  resultant  square 
and  the  given  square  in  Exercise  2? 

4.  Construct  a  square  whose  area  is  23  square  units;  37 
square  units;  8  square  units;  52  square  units.     Note  that 
A/52  =  A/4.13  =  2  Vl3.     Construct  Vl3  and  double  the 
result. 

5.  Find  the  altitude  of  an  equilateral  triangle  whose  side  is 
1  unit;  whose  side  is  6  units;  whose  side  is  25  units;  whose  side 
is  s  units.     See  illustration  which  follows. 

Solution.     Given  the  equilateral  triangle  ABC  with 
side  AB  1  unit  in  length. 

To  find  the  length  of  the  altitude  CD. 

A  D  B 

Analysis.  Since  ABC  is  an  equilateral  triangle,  the  altitude  CD  bi- 
sects the  base,  that  is,  AD  is  half  the  side  AB.  (Why?)  Then  ADC  is 
a  right-angled  triangle  with  hypotenuse  AC  one  unit  in  length,  and  side 
AD  half  a  unit  in  length.  Our  problem  is  to  find  the  length  of  side  CD. 

Let  h  =  the  number  of  units  in  CD. 

Then  (|)2  +  h*  =  I2,  by  the  Pythagorean  Theorem. 

7)2   —   1 

"      —    4>  _ 

h  =  ±  \  \/3,  number  of  units  in  altitude. 
Solve  each  of  the  other  parts  of  the  exercise  in  the  same  way. 
The  answer  to  the  last,  when  the  side  is  s  units  long,  is 

h  =  is  vj: 

The  student  should  fix  this  in  mind  as  a  formula.  It  will  be  needed 
many  times  in  the  future  work. 

6.  Using  the  formula  in  Exercise  5  find  the  altitude  of  the 
equilateral  triangles  whose  sides  are  7;  32;  280;  a  +  6;  4  m 


7.  If  the  altitude  of  an  equilateral  triangle  is  h,  find  the 
length  of  its  side. 

8.  Using  this  as  a  formula  find  the  length  of  the  side  of  the 
equilateral  triangle  whose  altitude  is  6;  11;  27;  3  m  —  2  n. 

9.  Find  the  areas  of  several  of  the  triangles  given  in  Exer- 
cises 5,  6,  7,  8. 

10.  Compute  the  area  of  a  regular  hexagon  the  length  of 
whose  side  is  1  unit.  * 


102 


PLANE  GEOMETRY 


[III,  §  116 


Suggestion.  In  the  adjoining  figure  AD  and  AE  are  diag- 
onals. FG  is  a  line  perpendicular  to  A  E. 
Let  the  student  show  that  Z  GFE  is 
60°,  Z  PEG  is  30°,  and  Z  AED  is  90°. 
Then  by  Art.  72  Ex.  5,  line  FG  is  J  ^#. 
From  this  find  the  length  of  diagonal 
A  E  and  then  A  D.  From  this  compute 
the  area  of  the  hexagon. 

11.  Compute  the  area  of  the  regular  hexagon  whose  side  is 
10  units  in  length. 

12.  Compute  the  area  of  a  regular  hexagon  whose  side  is  s 
units  in  length. 

13.  How  does  the  area  of  a  regular  hexagon  whose  side  is  s 
units  compare  with  the  area  of  an  equilateral  triangle  whose 
side  is  s  units?     (See  Ex.  14,  §  76.) 


14.  Compute  the  area  of  a  regular  hexa- 
gon by  means  of  the  adjoining  figure. 
Compare  with  Exercise  10. 


15.  In  the  adjoining  figure  A  BCD  is 
a  square  whose  side  AB  is  given.  It  is 
desired  to  cut  off  the  corners  so  as  to 
form  a  regular  octagon. 

Suggestion.    The  triangle  ARM  is  an 
isosceles  right  triangle.     We  must  find  M 
the  length  of  MR  in  terms  of  the  length 
of  AB  which  we  shall  assume  to  be  s. 
Let  d_  =  the  number  of  units  in  length  of  A  R. 

Then  d\/2_=  the  number  of  units  in  length  of  MR.  Why? 
Then  2  d  +  d\/2J=  the  number  of  units  in  length  of  AB.  Why? 
/.  2d+dV2=s.  Why? 


d  = 


V2 


-    (Rationalize  denominator  of  fraction.) 


Ill,  §  116] 


EQUAL  FIGURES 


103 


d  —  (1  —  JA/2)  s,  number  of  units  in  AR. 
dv/2~=  V(2  —  1)  s,     number  of  units  in  M#. 
Starting  with  a  square  construct  a  regular  octagon.     Do 
not  use  approximate  values.     Construct  V2  s  and  take  away 
s  from  it.     The  remainder  is  MR. 

16.  Suppose  that  the  figure  in  Ex.  15  is  a  regular  octagon 
the  length  of  whose  side  is  10  units.    By 

computing  the  areas  into  which  it  is 
divided  obtain  the  area  of  the  whole. 

17.  The  adjoining  figure  may  be  used 
as  a  design  for  tiling.     What  is  the 
shape  of  the  dark  blocks,  if  the  other 
blocks   are   regular  octagons?     Prove 
your  answer. 

18.  With  regular  octagonal  blocks  6  inches  on  the  side  it 
takes  10  blocks  one  way  and  12  blocks  the  other,  together  with 
a  sufficient  number  of  smaller  blocks  to  fill  in,  to  cover  a  floor. 
What  is  the  area  of  the  floor? 

19.  A  lady  made  56  patches  for  a  quilt.     Each  patch  was 
octagonal  in  shape,  was  made  of  5  squares  3  inches  in  dimen- 
sion when  finished,  and  4  triangular  pieces, 

as  shown  in  the  figure.  (Are  the  patches 
regular  octagons?)  The  blocks  were  put 
together  with  plain  muslin  as  shown  in  the 
figure.  A  strip  of  muslin  of  uniform  width 
forms  the  border.  The  quilt  when  finished 
is  2|  yards  long  and  2J  yards  wide.  How 
much  muslin  did  she  use?  Make  no  allowance  for  seams. 

20.  The  length  of  the  shorter  side  of  a  right  triangle  is  5J 
units  more  than  \  of  the  other.     The  hypotenuse  is  3  units 
less  than  twice  the  shorter  side.     What  is  the  length  of  each 
side  and  of  the  hypotenuse? 

21.  (a)  Give    three    sets    of    numbers,    other    than    the 
numbers  3,  4,  5,  or  multiples  of  them,  which  can  be  used  as 
the  sides  of  a  right  triangle. 


104 


PLANE  GEOMETRY 


[HI,  §  116 


21.  (6)  Show  that  ra2  —  n2,  2  mn,  ra2  +  n2  are  the  sides 
and  hypotenuse  of  a  right  triangle.     Show  this  by  making 
use  of  the  Pythagorean  theorem. 

22.  By  substituting  different  numbers  for  m  and  n  in  Exer- 
cise 21,  make  a  table  of  ten  sets  of  numbers  that  may  be  used 
as  the  sides  of  a  right  triangle. 

23.  From  the  table  of  Exercise  22  write  three  problems  con- 
cerning right  triangles. 

24.  In  the  adjoining  figure  let  a  be  the  length  of  one  side 
of  the  square.     Calculate  the  length  of  the 

oblique  lines  in  terms  of  a.  Calculate  the  area 
of  each  part  of  the  figure.  Construct  the  figure 
letting  a  =  2  inches.  Substitute  2  for  a  in 
your  answer  and  compare  results  with  those 
obtained  by  drawing. 

25.  This  is  a  design  for  a  parquetry  border. 
Construct  this  design,  making  your  drawing  4 
inches  wide.     Compute  the  area  of  each  part. 

26.  Construct  this  parquetry  design  for  a  border  based  on 
a  regular  octagon.     (Suggestion.     Lay  off  OB  equal  to  OA. 
Then  AB  is  the  side  of  a  regular  octagon.)     Prove  this  by 
proving  the  sides  equal  and  the  angles  equal » 


27.  Construct  parts  of 
borders,  four  times  as  wide 
as  in  these  illustrations. 


i 


Ill,  §  116] 


EQUAL  FIGURES 


105 


28.  Construct  a  parquetry  design  like 
the  illustration  so  that  the  dark  star  shall 
have  sides  one-half  inch  long.  Notice  that 
the  figure  can  be  built  up  from  a  system  of 
diagonal  lines.     Calculate  the  area  of  one 
of  the  stars  in  your  drawing. 

29.  Construct  this  parquetry  border  so 
that  the  lines  of  your  figure  shall  be  four 
times  as  long  as  those  of  the  illustration. 

30.  Construct  part  of  this  field  so 
that  the  lines  of  your  figure  shall  be 
four  times  as  long  as  in  the  illustra- 
tion.    Note  the  parallel  lines  as  a 
help  to  making  the  figure. 


31.  Construct  part  of  this  field 
using  sets  of  parallel  lines  placed  far 
enough  apart  to  make  the  hexagon 
come  out  one  inch  on  each  side. 


32.  Construct   the   following   figure,  taking   a  =  1  inch, 
Find  the  area  of  each  part. 


What  is  the  perimeter  of  one  of  the  black  triangles,  and 
of  one  of  the  parallelograms? 
What  are  their  areas? 


106 


PLANE  GEOMETRY 


,  § 


117.  Definition.  The  projection  of  a  point  on  a  line  is  the 
foot  of  the  perpendicular  drawn  from  the  point  to  the  line. 

From  what  two  Latin  words  is  the  word  projection  derived?  Do  you 
find  any  connection  between  the  original  meaning  and  the  meaning  here? 

If  we  draw  a  line-segment  A  B  and  another  unlimited  line 
XXi  in  the  same  plane,  and  if  we  imagine  every  point  on  A  B 
to  be  projected  upon  XX \  then  the  segment  of  XXi  which 
contains  all  of  these  projections  is  called  the  projection  of 
line-segment  A  B  upon  line 


B,    A, 


In  each  case  AiBi  is  the  projection  of  AB  upon 

118.  Theorem  X.  In  an  obtuse  angled  triangle,  the  square 
on  the  side  opposite  the  obtuse  angle  equals  the  sum  of  the  squares 
on  the  other  two  sides,  increased  by  twice  the  product  of  one  of 
those  sides  by  the  projection  of  the  other  side  on  the  line  of  that  side. 


Given  the  triangle  ABC  with  the  obtuse  angle  A CB. 

The  lengths  of  sides  BC,  CA,  and  AB  respectively  are  a, 
6,  c;  ai  the  projection  of  a  upon  the  line  of  side  6;  h  is  the  per- 
pendicular used  to  project  a  upon  line  of  b. 

To  prove  that  c2  =  a2  +  &2  +  2<tib. 

Analysis.  To  prove  this  we  find  in  our  figure  a  right  tri- 
angle of  which  c  is  the  hypotenuse,  and  then  by  means  of  the 


in,  §  119]  EQUAL  FIGURES  107 

Pythagorean  theorem  we  can  give  the  value  of  c  in  terms  of 
other  lines.  When  we  have  done  this,  we  shall  find  in  the 
equations  that  we  have  h2  and  ai2  which  are  not  called  for  in 
the  proposition,  while  a2  does  not  appear  in  the  equation.  If 
now  it  is  possible  to  substitute  a2  for  h2  +  ai2  in  the  equation, 
we  shall  have  proved  our  theorem. 

Proof.    A BD  is  a  right  triangle.  Why? 

c2  =  h2  +  Oi  +  b)2.  Why? 

=  h2  +  ai2  +  2a1b  +  b2.  Why? 

But          h2  +  «!2  =  a2.  Why? 

Substituting  a2  for  h2  +  «i2,  we  have 

c2  =  a2  +  2  aj)  +  62, 
State  the  theorem  proved. 

119.  Theorem  XI.  In  any  triangle  the  square  on  the  side 
opposite  an  acute  angle  is  equal  to  the  sum  of  the  squares  on  the 
other  two  sides,  diminished  by  twice  the  product  of  one  of  those 
sides  and  the  projection  of  the  other  side  upon  it. 


Give  the  proof  of  this  on  the  same  plan  as  in  the  preceding 
theorem,  noting  that  we  now  have  c2  =  h2  +  (b  —  fli)2. 

120.  Exercises.     If  an  engineer  wishes  to  find  the  distance 
across  a  swamp  which  is  impassable, 
he  may  use  the  following  plan: 

He  places  his  transit  at  the  point 
C  from  which  he  may  see  and  go  to 
both  ends  of  the  swamp.  By  sight- 
ing from  C  to  A  and  from  C  to  B,  cT 


108  PLANE  GEOMETRY  [in,  §  120 

he  can  find  angle  BCA.  (The  teacher  should  explain  the 
plan  on  which  the  transit  works.)  He  can  measure  the 
distance  from  C  to  A  and  from  C  to  B.  From  these  data  he 
can  find  the  distance  AB  by  making  use  of  theorems  IX, 
X,  and  XI,  according  to  the  size  of  Z  BCA. 

From  the  following  data  compute  this  distance,  taking 
CB  —  a  and  CA  =  6.  Draw  the  figures. 

l.o=    24  yds.,  b  =      7yds.,  Z  BCA  =    90°. 

2.  a  =    40yds.,  6  =    50yds.,  Z  BCA  =  150°. 

3.  a  =  100  yds.,  6  =  130yds.,  Z  BCA  =    30°. 

4.  a  =  120  yds.,  b  =    75yds.,  Z  BCA  =    60°. 

5.  a  =    60  yds.,  b  =  120yds.,  Z  BCA  =  120°. 

It  is  shown  in  physics  that  the  resultant  of  two  forces  act- 
ing at  the  same  instant  upon  an  object  is  the  same  as  though 
one  of  the  forces  acted  and  when 
this  action  finished,  the  second 
force  acted.  In  the  adjacent  fig- 
ure, if  two  mallets  hit  a  ball  at  the 
same  instant,  the  first  of  which  °  a 

would  send  the  ball  the  distance  a  feet  in  the  direction  in- 
dicated, and  the  second  the  distance  b  in  the  direction 
indicated,  the  resultant  distance  would  be  the  distance  c  in 
the  direction  indicated. 

Also  the  velocity  with  which  an  object  will  move  when 
acted  upon  by  two  forces  can  be  obtained  if  we  know  the 
velocity  due  to  each  of  the  forces,  and  the  angle  at  which 
they  act  with  each  other. 

From  drawings  on  cross  section  paper  measure  the  resultant 
force  or  velocity  in  the  following  cases;  Z  ab  means  the  angle 
between  a  and  6. 

6.  a  =    25,  b  =    50,  Z  ab  =    54°. 

7.  a  =    70,  b  =    45,  Z  ab  =  170°. 

8.  a  =  100,  b  =  120,  Z  ab  =    93°. 


ill,  §  120]  EQUAL  FIGURES  109 

In  the  following,  after  drawing,  compute  the  resultant  force 
or  velocity  and  compare  answers. 
9.  a  =  100,  b  =  20,  Z  ab  =    30°. 

10.  a  =    10,  b  =  50,  Z  ab  =    90°. 

11.  a  =  100,  b  =  20,  Z  ab  =    90°. 

12.  a  =  100,  6  =20,  Z  ab  =  150°. 

13.  a  =    40,  b  =  40,  Z  ab  =  120°. 

(For  illustration  of  solution  of  following  exercises,  see  First 
Course,  pages  194, 195.) 

14.  Two  forces  are  acting  at  right  angles  to  one  another. 
The  smaller  force  is  5  pounds  less  than  J  the  larger  force,  and 
the  resultant  force  is  9  pounds  more  than  the  larger.     What  is 
the  number  of  pounds  in  the  resultant  force?     Draw. 

15.  If  rain-drops  which  are  falling  vertically  at  the  rate  of 
120  feet  per  second  enter  an  air  current  which  is  moving  hori- 
zontally at  50  feet  per  second,  what  will  be  the  resultant  veloc- 
ity of  the  rain-drops?     Make  a  drawing  showing  their  direc- 
tion. 

16.  From  a  moving  body  a  particle  is  thrown  at  right  angles 
to  the  direction  in  which  the  body  is  moving.    The  velocity 
of  the  particle  as  it  leaves  the  body  is  4  miles  per  second  more 
than  the  velocity  of  the  body,  while  the  velocity  at  which  it 
is  thrown  from  the  body  is  f  of  a  mile  more  per  second  than  J 
of  the  velocity  with  which  it  leaves  the  body.     What  is  the 
velocity  of  the  particle?     Draw. 

17.  A  ship  is  sailing  due  southeast  at  the  rate  of  llj/g  miles 
an  hour  in  an  ocean  current  which  is  flowing  due  north-east 
at  the  rate  of  2J/2  miles  an  hour.     What  is  the  resultant 
speed  of  the  ship?     Show  by  a  diagram  the  direction  in  which 
the  ship  actually  moves. 

18«  How  much  sod  will  it  take  to  resod  a  corner  of  a  lawn, 
which  is  in  the  shape  of  a  right-angled  triangle,  if  the  length 
of  the  longer  side  is  3  feet  more  than  that  of  the  shorter  side, 
and  the  length  of  the  shorter  side  is  4  feet  more  than  J  of  the 
length  of  the  hypotenuse? 


110  PLANE  GEOMETRY  IIII,  §  121 

121.  Summary  of  Chapter  III. 

Equal  or  Equivalent  Figures. 

Definition  of  Equal  Figures,  Addition  and  Subtraction  of  Areas. 

Definition  of  Base  and  Altitude  of  Parallelogram,  Triangle,  Trape- 
zoid. 

Theorem  I.  Parallelograms  on  the  same  or  equal  bases  and  between 
the  same  parallel  lines  are  equal. 

Corollary  1.  Parallelograms  having  equal  bases  and  altitudes  are 
equal. 

Corollary  2.  A  parallelogram  is  equal  to  a  rectangle  which  has  an 
equal  base  and  stands  between  the  same  parallel  lines. 

Theorem  II.  Triangles  which  stand  on  the  same  or  equal  bases  and 
between  the  same  parallel  lines  are  equal. 

Corollary  1.  Triangles  having  equal  bases  and  equal  altitudes  are 
equal. 

Corollary  2.  A  triangle  equals  half  a  parallelogram  on  the  same  or 
equal  bases  and  between  the  same  parallel  lines. 

Theorem  III.  A  trapezoid  is  equal  to  half  a  rectangle  whose  base  is 
the  sum  of  the  two  parallel  sides,  and  whose  altitude  is  the  altitude  of 
the  trapezoid. 

Assumption.  The  area  of  a  rectangle  equals  the  product  of  its  base 
by  its  altitude. 

Corollary  1.  The  area  of  a  parallelogram  equals  the  product  of  its 
base  by  its  altitude. 

Corollary  2.  The  area  of  a  triangle  equals  the  product  of  half  its 
base  by  its  altitude. 

Corollary  3.  The  area  of  a  square  is  equal  to  the  square  of  one  of 
its  sides. 

Corollary  4>     The  area  of  a  trapezoid  equals  the  product  of  half  the 
sum  of  its  bases  by  its  altitude. 
Sums  and  Differences  of  Areas. 

Theorem  IV.  The  rectangle  of  two  given  lines  equals  the  sum  of 
the  rectangles  contained  by  one  of  them  and  the  several  segments  into 
which  the  other  is  divided. 

Theorem  V.  The  square  on  the  sum  of  two  lines  equals  the  sum  of 
the  squares  on  those  lines  plus  twice  their  rectangle. 

Theorem  VI.  The  rectangle  of  a  given  line  and  the  difference  of  two 
other  lines  equals  the  difference  of  the  rectangles  of  the  first  line  and 
each  of  the  others. 

Theorem  VII.  The  square  on  the  difference  of  two  lines  is  equal  to 
the  sum  of  the  squares  on  those  lines  minus  twice  their  rectangle. 


in,  §  121]  EQUAL  FIGURES  111 

Theorem  VIII.  The  difference  of  the  squares  on  two  lines  is  equal 
to  the  rectangle  of  the  sum  and  difference  of  those  lines. 

Theorem  IX.  The  square  on  the  hypotenuse  of  a  right  triangle  is 
equal  to  the  sum  of  the  squares  on  the  other  two  sides. 

Problem  I.     To  construct  a  right  triangle  when  two  sides  are  given. 

Problem  II.  To  construct  a  right  triangle  when  a  side  and  hypot- 
enuse are  given. 

Problem  III.  To  construct  a  square  that  will  be  the  sum  of  two 
given  squares. 

Problem  IV.     To  construct  the  square  root  of  a  number. 
Definition.     Projections. 

Theorem  X.  In  an  obtuse  angled  triangle  the  square  on  the  side 
opposite  the  obtuse  angle  is  equal  to  the  sum  of  the  squares  on  the 
other  two  sides,  plus  twice  the  product  of  one  of  these  sides  and  the  pro- 
jection of  the  other  side  on  the  line  of  that  side. 

Theorem  XI.  In  any  triangle  the  square  on  the  side  opposite  an  acute 
angle  is  equal  to  the  sum  of  the  squares  on  the  other  two  sides,  minus 
twice  the  product  of  one  of  these  sides  by  the  projection  of  the  other 
side  upon  it. 

1.  Of  what  subject  matter  does  Chapter  III  treat? 

2.  Can  figures  be  equal  without  being  congruent?     Explain. 

3.  Name  various  figures  that  you  know  to  be  equal,  as  brought  out 
in  this  chapter.     Name  figures  that  you  know  to  have  half  the  area  of 
other  figures. 

4.  To  what  subject  in  algebra  is  part  of  this  chapter  closely  related? 

5.  What  famous  theorem  is  given  in  this  chapter?     By  what  name 
is  it  known  and  why? 

6.  Go  through  the  list  of  theorems  (page  reference).     Do  you  find 
theorems  that  have  been  proved   without  the  use  of  theorems   of 
Chapter  II? 

The  theorems  of  this  chapter  are  selected  from  Book  II  of  Euclid's 
Geometry. 


CHAPTER  IV 

PART   I— CIRCLES.     CENTRAL   ANGLES.     CHORDS. 
PART  II— INSCRIBED  ANGLES. 

PART  I— CIRCLES.    CENTRAL  ANGLES.    CHORDS. 

122.  Definitions.     See  also  Articles  41  and  42. 

A  circle  is  the  portion  of  a 
plane  bounded  by  a  curved 
line,  all  points  of  which  are 
equidistant  from  a  point 
within  called  the  center. 

The  curved  line  is  called  the  circumference  of  the  circle. 

Note.  This  is  true  in  the  elementary  geometry,  but  as  you  go  farther  in  your  work, 
you  will  find  often  that  the  curved  line  is  called  the  circle. 

A  straight  line  terminated  by  the  center  and  the  circum- 
ference is  called  a  radius  of  the  circle. 

A  straight  line  passing  through  the  center  of  the  circle  and 
terminated  by  the  circumference  is  called  a  diameter  of  the 
circle. 

A  line  cutting  the  circumference  is  called  a  secant. 

The  segment  of  a  secant  cut  off  by  a  circumference  is 
called  a  chord. 

Equal  circles  are  those  the  diameters  of  which  are  equal, 
or  the  radii  of  which  are  equal. 

Exercises.  1.  A  circle  passes  through  the  four  corners 
of  a  square.  Show  that  a  diagonal  of  the  square  is  a  diameter 
of  the  circle.  What  is  the  diameter  of  the  circle  if  the  side 
of  the  square  is  a  units? 

112 


IV,  §123]  CIRCLES.     CENTRAL  ANGLES  113 

2.  Draw  a  circle  on  a  given  line-segment  as  diameter. 

123.  Postulates  for  the  Circle.     The  following  assumptions 
are  made: 

Post.  VII.  A  circumference  may  be  drawn  with  any  point  as 
center  and  with  any  given  line-segment  as  radius. 

Post.  VIII.  A  point  is  within,  on,  or  with- 
out a  circumference  according  as  its  distance 
from  the  center  is  less  than,  equal  to,  or  greater 
than  the  radius. 

Exercise.  Draw  a  square  with  sides  2  inches  long,  and 
draw  circles  on  each  side  of  the  square  as  diameters.  Show 
that  these  circles  are  equal  and  that  they  will  all  pass  through 
the  center  of  the  square. 

124.  Definitions.     An  arc  is  any  part  of  a  circumference. 
One-half  of  a  circumference  is  called  a  semicircumf erence. 

A  sector  is  the  portion  of  a  plane  bounded  by 
two  radii  of  a  circle,  and  the  arc  which  they 
intercept. 

One-half  of  a  circle  is  called  a  semicircle. 

A  central  angle  is  an  angle  whose  arms  are  radii. 

We  use  the  expression:  A  central  angle  stands  on  an  arc, 
or  intercepts  the  arc.  The  arc  subtends  the  central  angle. 


A  chord  subtends  two  arcs.     The  smaller  arc  is  called  the 
minor  arc,  and  the  larger  arc  is  called  the  major  arc. 


114  PLANE  GEOMETRY  [IV,  §  125 

125.  Theorem  I.     In  the  same  or  equal  circles,  if  two  central 
angles  are  equal,  their  intercepted  arcs  are  equal. 


Given  the  equal  circles  whose  centers  are  0  and  Oi  with 
the  equal  central  angles  AOB  and  AiOiB^  intercepting  the 
arcs  AB  and  A^Bi  respectively. 

To  prove  that  arc  AB  is  equal  to  the  arc  AiB\. 

Analysis.  Since  this  is  the  first  proposition  that  we  have 
to  prove  with  reference  to  circles,  we  shall  use  the  method  of 
superposition.  Placing  the  circle  whose  center  is  0  on  the 
circle  whose  center  is  Oi,  we  must  show  that  every  point  in 
arc  AB  falls  upon  a  corresponding  point  in  arc  AiB±. 

Proof.  Place  the  circle  whose  center  is  0  on  the  circle 
whose  center  is  Oi,  so  that  the  Z  A  OB  coincides  with 
Z  AtOiBi.  Then  A  falls  on  AI.  Why? 

Also  B  falls  on  BI.     Why? 

Then  every  point  in  arc  A  B  falls  upon  a  corresponding 
point  in  arc  AiBi.  Why? 

State  proposition  proved.     State  the  converse  theorem. 

Corollary  1.  A  diameter  divides  a  circumference  into  two 
equal  parts.  Explain. 

Corollary  2.  In  the  same  or  equal  circles  sectors  which  have 
equal  angles  are  equal.  Why? 

Corollary  3.  A  diameter  divides  the  circle  into  equal  parts. 
Why? 


IV,  §126]  CIRCLES.    CENTRAL  ANGLES  115 

126.  Theorem  II.    State  the  converse  of  Theorem  I.    (See 
Summary.)     Prove  by  placing  the  two  circles  together,  so 
that  the  centers  fall  together,  and  the  given  equal  arcs  coin- 
cide.    Show  that  the  angles  coincide  and  are  therefore  equal. 

Corollary  1.     State    and   prove    converse    of    Theorem   I, 
Corollary  2. 

127.  Theorem  III.     In  the  same  or  equal  circles  if  two 
central  angles  are  unequal,  the  arcs  which  they  intercept  are 
unequal,  and  the  greater  angle  intercepts  the  greater  arc. 


Given  the  equal  circles  whose  centers  are  0  and  Oi  with 
the  unequal  central  angles  AOB  and  AiOiBi  intercepting  the 
arcs  A B  and  AiBi  respectively,  angle  AOB  being  less  than 
angle  AiOiBi. 

To  prove  that  arc  AB  is  less  than  arc  AiBi. 

Analysis.  We  shall  prove  this  by  placing  the  circle  whose 
center  is  0  on  the  circle  whose  center  is  Oi,  with  center  0  on 
Oi,  and  the  arm  OA  on  arm  OiAi.  From  this  we  can  show 
that  the  initial  ends  of  the  arcs  AB  and  AiBi  coincide,  and 
that  the  final  ends  do  not  coincide. 

Proof.  Placing  the  circle  with  center  0  on  the  circle  with 
center  Oi,  0  being  placed  on  Oi,  and  OA  on  OiAi,  point  A  will 
fall  on  point  A\.  Why? 

OB  will  fall  between  OiAi  and  OiBi,  because  Z  AOB  < 
Z  AiOiBi. 

.'.     B  will  fall  between  AI  and  ft. 

/.     arc  AB  <  arc  AiBi. 

State  the  theorem  proved.     State  the  converse  theorem. 


116  PLANE  GEOMETRY  [IV,  §  128 

Theorem  IV.  State  and  prove  the  converse  of  Theorem 
III.  (See  Summary.) 

Theorems  I  and  III  are  usually  stated  together  as  one 
theorem. 

128.  Measurement  of  Arcs.     Since  equal  angles  intercept 
equal  arcs,  each  of  the  360  equal  angles  into  which  we  divided 
a  perigon  in  order  to  establish  the  degree  unit,  will  intercept 
one  three  hundred  and  sixtieth  of  a  circumference  described 
about  the  vertex  as  a  center.     From  this  fact  we  speak  of  one 
three  hundred  and  sixtieth  of  a  circumference  as  an  arc  of  one 
degree.     Thus  a  unit  for  measuring  an  arc  is  established.    An 
arc  of  20  degrees  is  intercepted  by  a  central  angle  of  20  degrees. 

You  will  frequently  see  the  expression  that  "an  arc  mea- 
sures an  angle."  We  shall  use  this  expression  in  our  work 
and  it  is  to  be  understood  that  the  unit  for  measurement  of 
arc  is  derived  as  above  from  the  unit  of  angle. 

129.  Theorems  I,  III.     In  the  same  circle  or  equal  circles 
if  two  central  angles  are  equal,  the  intercepted  arcs  are  equal, 
and  of  two  unequal  central  angles  the  greater  angle  intercepts  the 
greater  arc. 

This  theorem  suggests  the  dependence  of  one  magnitude 
on  another,  in  this  case  the  dependence  of  arc  on  central 
angle,  so  we  will  now  consider  this  subject.  At  this  time  it 
will  be  well  for  the  student  to  review  First  Course,  pages  206 
to  214. 

130.  Definition.     One  quantity  is  said  to  be  a  function  of 
another  quantity  when  it  depends  on  that  quantity  for  its 
value. 

From  the  above  theorem  we  may  state  that  the  length  of  the 
arc  depends  on  the  size  of  the  central  angle,  that  is, 

The  length  of  the  arc  is  a  function  of  the  size  of  the  central 
angle,  if  the  circle  remains  the  same  size. 


IV,  §  131]  CIRCLES.     CENTRAL  ANGLES  117 

Other  Examples: 

In  general,  the  size  of  a  tree  depends  on  the  number  of 
years  that  it  has  been  growing,  that  is, 

The  size  of  a  tree  is  a  function  of  the  number  of  years  of 
growth. 

The  rise  and  fall  of  the  mercury  in  the  thermometer  de- 
pends on  the  temperature,  that  is, 

The  height  of  the  mercury  in  a  thermometer  is  a  function 
of  the  temperature. 

The  weight  of  a  ball  of  given  material  depends  on  the 
volume,  that  is, 

The  weight  of  a  ball  is  a  function  of  the  volume.  Let  the 
student  give  other  illustrations  of  this  kind. 

Drawing  illustrations  from  our  past  work  in  geometry,  we 
have: 

The  area  of  a  rectangle  is  a  function  of  the  altitude  if  the 
base  remains  the  same,  or 

The  area  of  a  rectangle  is  a  function  of  the  base  if  the  alti- 
tude remains  the  same. 

Let  the  student  go  through  the  theorems  which  have  been 
studied  and  decide  upon  those  that  involve  the  dependency 
of  one  magnitude  upon  another,  and  re-state  some  of  them, 
using  the  word  function. 

131.  Algebraic  Expression  of  the  Function  Idea. 

In  the  illustration  of  the  rectangle 

a  =  bh 

is  the  algebraic  expression  for  the  area  of  a  rectangle. 
Then  a  =f(h),  (read  "a  is  a  function  ofh") 

where  /  (ti)  =  bh, 

is  the  algebraic  way  of  stating  that  if  the  base  remains  the 
same,  the  area  of  a  rectangle  is  a  function  of  its  altitude. 
Also  a  =f(b),  (r *ead  "a  is  a  function  of b") 

where  /  (b)  =  bh, 


118  PLANE  GEOMETRY  [IV,  §  132 

is  the  algebraic  way  of  stating  that,  if  the  altitude  remains  the 
same,  the  area  of  a  rectangle  depends  on  its  base. 
Finally  a=f(b,h)    (read  "ais  a  function  ofb  and  /i") 

is  the  algebraic  way  of  stating  that  if  the  base  and  altitude 
both  change,  the  area  of  the  rectangle  depends  on  them  both. 

132.  Exercises. 

1.  Make  drawings  illustrating  each  of  the  above  state- 
ments.    (Cross-section  paper  should  be  used.) 

2.  As  shown  in  the  illustration  above,  give  algebraic  state- 
ments for  the  area  of  a  parallelogram. 

3.  Make  similar  statements  for  the  changes  in  the  area  of  a 
triangle. 

4.  If  r,  x,  y  are  the  hypotenuse  and  the  sides  of  a  right  tri- 
angle, we  have  r2  —  x2  +  2/2- 

Tell  what  the  following  expressions  mean: 
r -/(*), 
r  =/(»), 
r=f(x,y). 

It  will  be  noticed  in  the  above  exercises  there  are  two  kinds 
of  quantities,  namely,  variables  and  constants.  Of  the  vari- 
able numbers  there  are  two  kinds,  the  independent  variables 
or  those  which  we  change  at  will,  and  the  dependent  variables 
or  those  which  change  because  another  number  changes.  The 
dependent  variables  are  functions  of  the  independent  vari- 
ables. 

In  the  theorems  studied  in  the  future  note  carefully  these 
quantities. 

Theorems  II  and  IV  may  be  stated  in  one  theorem. 

133.  Theorems  II,  IV.     In  the  same  circle  or  equal  circlesf 
if  arcs  are  equal,  they  are  intercepted  by  equal  central  angles,  and 
of  two  unequal  arcs  the  greater  arc  is  intercepted  by  the  central 
greater  angle. 

In  this  theorem,  what  is  the  independent  variable? 
What  is  the  dependent  variable?  What  is  the  constant? 


IV,  §134]  CIRCLES.     CENTRAL  ANGLES  119 

Use  the  word  function  to  state  the  relation  between  the  in- 
dependent and  dependent  variables. 

Explain  how  the  independent  variable  of  Theorem  III 
becomes  the  dependent  variable  in  Theorem  IV,  and  vice 
versa. 

THEOREMS  ON  ARCS  AND  CHORDS 

134.  Theorem  V.  In  the  same  or  equal  circles,  if  two  arcs 
are  equal  they  are  subtended  by  equal  chords. 


Given  the  equal  circles  whose  centers  are  0  and  Oi  with 
the  arcs  A B  and  AiBi  equal,  subtended  by  the  chords  AB 
and  AiBi  respectively. 

To  prove  that  chord  AB  is  equal  to  chord  AiBi. 

Analysis.  We  can  prove  chord  AB  and  chord  AiBi  equal 
if  we  draw  the  radii  OA,  OB,  OiAi,  O^Bi,  and  prove  the  tri- 
angles thus  formed  congruent.  We  can  prove  the  triangles 
congruent,  if  we  can  prove  the  A  AOB,  AiO\B±  equal,  since 
we  know  the  radii  are  equal.  Z  AOB  equals  Z  AiOiBi  if 
arcs  AB  and  AiBi  are  equal.  But  these  are  equal  by 
hypothesis. 

Proof.     Make  this  proof  complete. 

135.  Theorem  VI.  State  the  converse  of  Theorem  V. 
(See  Summary.) 

Prove  by  using  congruent  triangles. 


120  PLANE  GEOMETRY  [IV,  §  136 

136.  Theorem  VII.  In  the  same  or  equal  circles,  if  two 
minor  arcs  are1  unequal,  the  chords  by  which  they  are  subtended 
are  unequal,  and  the  greater  arc  is  subtended  by  the  greater  chord. 


Given  the  equal  circles  whose  centers  are  0  and  0\y  with 
minor  arc  A B  greater  than  minor  arc  A\Bi. 

To  prove  that  chord  AB  is  greater  than  AiBi. 

Analysis.  To  prove  this  we  form  two  triangles  OAB  and 
OiAiBi  by  drawing  radii  OA,  OB,  OiAi,  OiBi.  We  can  prove 
line  AB  greater  than  AiBi  if  we  can  show  that  the  sides  OA 
and  OB  are  equal  respectively  to  OiAi  and  OiBi  and  that  the 
included  angle  AOB  is  greater  than  the  included  angle  AiOiBi. 

Proof.  arc  AB  >  arc  A^.  Why? 

Z  AOB  >  Z  AiOiBi.         Art.  133. 
OA  =  OiAi, 

OB  =  Oifii.  Why? 

side  AB  >  side  AiBi.          Art.  88. 
State  theorem  proved. 
State  converse  of  this  theorem. 

Exercise.  A  circumference  is  divided  into  four  equal 
arcs,  and  their  chords  are  drawn. 

(a)  Show  that  the  sides  of  the  quadrilateral  are  equal. 

(b)  Show   that   the   diagonals   of   the    quadrilateral   are 

diameters. 

(c)  Show  that  the  quadrilateral  is  a  parallelogram. 

(d)  Show  that  the  quadrilateral  is  a  rectangle. 


IV,  §  137] 


CIRCLES.     CENTRAL  ANGLES 


121 


137.  Theorem  VIII.  In  the  same  or  equal  circles  if  two 
chords  are  unequal,  they  subtend  unequal  minor  arcs,  and  the 
greater  chord  subtends  the  greater  arc. 


Given  the  equal  circles  whose  centers  are  0  and  Oi,  with 
chord  AB  greater  than  AiBlt 

To  prove  that  arc  AB  is  greater  than  arc  AiBi. 

Analysis.  To  prove  this  we  form  two  triangles  OAB  and 
OiAiBi  by  drawing  radii  OA,  OB,  OiAi  and  OiBi.  To  prove 
arc  AB  greater  than  arc  AiBi,  we  shall  prove  Z  AOB  greater 
than  Z  AiOiBi.  We  can  do  this  by  applying  the  theorem 
of  Art.  89,  page  75,  to  the  two  triangles. 

Proof.  side  AB  >  A^.  Why? 

side  OA  =  O^A^ 

side  OB  =  Oi#i.  Why? 

Z  AOB  >  Z  AiOiBi.  Art.  89. 

arc  AB  >  arc  AiBi.  Art.  128. 

138.  Theorem  IX.  A  diameter  which  is  perpendicular  to  a 
chord  bisects  the  chord  and  its  subtended  arc. 


Prove  by  drawing  in  radii  and  showing  congruent  triangles. 


122 


PLANE  GEOMETRY 


[IV,  §  139 


139.  Theorem  X.     //  a  diameter  bisects  a  chord,  it  is  per- 
pendicular to  the  chord. 

Prove  by  drawing  radii  and  proving  congruent  triangles. 

140.  Theorem  XI.     The  perpendicular  bisector  of  a  chord 
passes  through  the  center  of  the  circle,  and  bisects  the  subtended 
arc. 

Q 


P  T 


Line  PQ 


Given  the  circle  with  center  0  and  chord  AB. 
is  perpendicular  to  A  B  at  the  mid-point  M. 

To  prove  that  PQ  passes  through  center  O. 

Analysis.  We  shall  suppose  PQ  does  not  pass  through 
center  0.  We  shall  draw  the  diameter  H  T  through  the  mid- 
point M .  We  can  prove  that  PQ  passes  through  0  if  we  can 
show  that  it  coincides  with  H  T.  We  can  show  that  it  coin- 
cides with  HT  if  we  can  show  that  it  makes  the  same  angle 
with  A  B  that  H  T  does.  We  can  show  that  both  PQ  and  H  T 
make  right  angles  with  AB. 

Proof.    HT_LABatM. 

Z  BMH  is  a  right  angle. 

PQ_LA5atM.     Why? 

Z  BMP  is  a  right  angle. 

PQ  and  H  T  coincide. 

PQ  passes  through  center  0. 
State  proposition  proved. 


Why? 
Why? 

(Hypothesis.) 

Why? 
Why? 


IV,  §  141]  INSCRIBED  ANGLES  123 

Corollary.     In  general,  two  chords  can  not  bisect  each  other. 
What  is  the  exception? 

Exercises. 

1.  Through  a  given  point  in  a  given  circle,  draw  a  chord 
Which  shall  be  bisected  at  the  point. 

2.  The  line  joining  the  mid-points  of  two  chords  passes 
through  the  center.     Show  that  the  chords  are  parallel. 


THEOREMS  ON  CHORDS  AND  DISTANCES  FROM 
THE  CENTER 

141.  Theorem  XII.     In  the  same  circle  or  equal  circles  equal 
chords  are  equidistant  from  the  center. 

Note.    By  distance  from  the  center  we  mean  the  perpen- 
dicular distance. 


Given  the  equal  circles  whose  centers  are  0  and  Oi  with 
the  equal  chords  AB  and  AiBi  at  the  distances  OH  and  0\Hi 
respectively. 

To  prove  OH  equal  to  OiHi. 

Prove  by  drawing  radii  OA  and  OiAi  and  showing  con- 
gruent triangles. 

142.  Theorem  XIII.  State  and  prove  the  converse  of 
Theorem  XII.  (See  Summary.) 


124  PLANE  GEOMETRY  [iv,  §  143 

143.  Theorem  XIV.  In  the  same  or  equal  circles,  if  two 
chords  are  unequal,  they  are  unequally  distant  from  the  center 
and  the  greater  chord  is  the  less  distant. 


Given  the  circle  whose  center  is  0,  with  chord  AB  greater 
than  chord  CD. 

To  prove  that  AB  lies  nearer  O  than  does  CD. 

Analysis.  Suppose  chord  AE  drawn  equal  to  chord  CD. 
Let  OH  and  OF  be  _k  from  0  to  AB  and  AE  respectively. 
Then  OH  and  OF  are  sides  of  a  triangle,  and  we  can  prove 
OH  <  OF  if  we  can  prove  Z  OFH  <  Z  OHF.  (See  Art.  70.) 
But  these  angles  are  complements  of  Z  AFH  and  Z  AHF 
respectively  in  A  AHF.  From  what  you  know  about  the 
lengths  of  AF  and  AH,  can  you  prove  Z  AFH  >  £AHF1 
(See  Art.  53.)  Then  what  follows  regarding  Z  OFH  and 
Z  OHFt  Then  what  do  you  conclude  about  OH  as  com- 
pared with  OF. 

Proof.     Let  the  student  give  the  proof  in  full. 

Algebraic  Proof.  Given  the  circle  whose  center  is  0  with 
the  chord  A  B  at  the  distance  OH  from  the  center.  (Figure 
below.) 


IV,  §  144]  INSCRIBED  ANGLES  125 

To  prove  that  if  chord  AB  is  to  become  longer,  it  must  move 
nearer  to  the  center. 

Let  d  =  the  number  of  units  in  OH; 

c  =  the  number  of  units  in  AH', 

r  =  the  number  of  units  in  OA. 

A  A  gQ  is  a  right  triangle.     Why? 
Therefore       d  =  Vr2  -  c2.     Why? 

We  may  regard  c  as  an  independent  variable  in  this  expres- 
sion. Then  d  will  become  the  dependent  variable.  In  other 
words  we  shall  regard  d  =  f  (c) . 

The  quantity  r2  —  c2  will  become  smaller  as  the  value  of  c 
becomes  larger,  because  the  less  remainder  is  obtained  by  sub- 
tracting the  greater  magnitude.  Therefore  d  becomes  smaller 
as  c  becomes  larger. 

State  the  theorem  proved. 

144.  Theorem  XV.     State  the  converse  of  Theorem  XIV. 
(See  Summary.) 

Give  proofs  similar  to  those  of  Theorem  XIV. 
Corollary.     The  diameter  is  the  longest  chord  in  a  circle. 

145.  Exercises.     Geometric  and  Algebraic. 

1.  If  through  a  point  on  the  circumference  of  a  circle, 
chords  are  drawn  which  make  equal  angles  with  the  radius 
drawn  to  that  point,  the  chord  will  be  equal. 

Suggestion.  Prove  that  they  are  equally  distant  from  the 
center. 

2.  A  chord  is  drawn  in  a  circle;  another  chord  is  drawn 
through  its  mid-point;  another  chord  is  drawn  through  the 
mid-point  of  the  last,  and  so  on.     Prove  that  these  chords 
continue  to  grow  longer.     Point  out  the  independent  and 
dependent  variables,  and  the  constant. 

Suggestion.  Prove  that  the  distance  from  the  center  is 
less  for  each  successive  chord. 


126 


PLANE  GEOMETRY 


[IV,  §  145 


3.  A  chord  MN  is  drawn  in  a  circle.     Chords  XY  are  drawn 
so  that  they  continually  have  their  mid-points  on  the  chord 
MN.     What  is  the  length  of  the  longest  and  of  the  shortest 
chord  XY. 

Point  out  the  independent,  the  dependent  variables  and  the 
constant. 

4.  A  circle  whose  center  is  on  the  bisector  of  an  angle  cuts 
off  equal  chords,  if  any,  from  the  arms  of  the  angle. 

6.  If  two  equal  chords  intersect  within  a  circle,  one  set  of 
lines  joining  their  ends  are  equal. 

6.  Of  all  the  chords  that  can  be  drawn  through  a  point 
within  a  circle  the  one  perpendicular  to  the  diameter  through 
that  point  is  the  shortest. 

7.  If  from  the  ends  of  the  diameter  of  a  circle  lines  are 
drawn  perpendicular  to  a  secant  of  the  circle,  prove  that  the 
segments  cut  between  the  circumference  and  the  feet  of  the 
perpendiculars  are  equal. 

Make  two  figures,  one  in  which  the  secant  cuts  the  diam- 
eter, and  the  other  in  which  it  does  not  cut  the  diameter. 

Suggestion.  Draw  a  line  from  the  center  of  the  circle  per- 
pendicular to  the  secant. 

8.  Figure  ABC  is  a  Gothic  arch.     Arcs  AC  and  BC  are 
drawn  from  A  and  B  as  centers  and  with 

radius  equal  to  line  A  B.      Construct  such 
an  arch. 

Show  that  each  arc  is  one-sixth  of  a  cir- 
cumference. 

Find  the  height  of  the  arch  if  line  AB  = 
60  inches. 

If  the  height  of  the  arch  is  to  be  30 
inches,  how  wide  must  it  be? 

9.  Construct    a    figure    in  which  two 
smaller  and  equal  Gothic  arches  are  con- 
tained in  a  larger  arch.    Find  the  height  of 
each  arch  if  the  large  arch  is  4  inches  wide. 


IV,  §  146]  INSCRIBED  ANGLES  127 

146.  Theorem  XVI.  Of  all  lines  passing  through  a  point 
on  a  circumference,  the  perpendicular  to  the  radius  drawn  to 
that  point  is  the  only  one  that  does  not  meet  the  circumference 
again. 


Given  the  circle  whose  center  is  0,  with  point  P  on  the 
circumference. 

Through  P  line  MN  is  drawn  perpendicular  to  the  radius 
OP.  PQ  is  drawn  oblique  to  OP. 

To  prove  that  MN  does  not  meet  the  circumference  again  and 
that  PQ  does  meet  it  again. 

Analysis.  In  order  to  prove  that  MN  does  not  meet  the 
circumference  again,  we  prove  that  every  point  on  MN  other 
than  P  is  more  distant  from  0  than  is  P.  In  order  to  prove 
that  PQ  cuts  the  circumference  again,  we  must  prove 
that  there  exists  at  least  one  point  on  PQ  at  a  less  distance 
than  OP  from  0. 

Proof.  Select  any  point  other  than  P  on  MN,  say  point 
R.  Join  0  and  R. 

OR  >  OP.     Why? 

Therefore  MN  does  not  meet  the  circumference  again. 

Since  OP  is  not  perpendicular  to  PQ,  drop  a  perpendicular 
to  PQ  from  0.  Call  it  OH. 

OIK  OP.    Why? 

Therefore  there  is  a  point  H  on  PQ  that  is  nearer  0  than  is 
P,  and  hence  lies  within  the  circle.  Art.  123.  Hence  the 
PQ  cuts  the  circumference  again.  Why? 


128  PLANE  GEOMETRY  [IV,  §  147 

Exercises. 

1.  In  the  figure  on  page  127,  suppose  a  point  T  marked 
on  PN,  so  that  PT  =  OP.     At  T  draw  a  line  perpendicular 
to  PN.     Show  that   it  will  meet  the  circle  in  only  one 
point. 

2.  Show  how  to  draw  a  circle  with  a  given  radius  and 
which  shall  be  met  by  a  given  line  in  only  one  point.     Draw 
several  such  circles. 

3.  Draw  a  circle  with  a  radius  of  1.5  inches,  and  let  P  be 
a  point  2.5  inches  from  the  center.     With  P  as  center  and  a 
radius  2  inches  draw  an  arc  cutting  the  circle  in  Q.     Will 
line  PQ  cut  the  circle  once,  twice,  or  not  at  all?     Prove. 

147.  Definition. 

A  tangent  to  a  circle  is  an  unlimited  straight  line  which 
meets  the  circumference  in  only  one  point.  This  point  is  called 
the  point  of  tangency,  or  the  point  of  contact. 

When  we  speak  of  a  tangent  from  a  point 
to  a  circle,  we  mean  the  segment  of  the 
tangent  between  the  point  and  the  point  of 
contact. 

148.  Theorem  XVII.     One  and  only  one  tangent  can  be 
drawn  to  a  circle  at  a  point  on  its  circumference. 

Suggestion.  Draw  the  radius  to  the  point  and  try  to  draw 
two  tangents  at  the  point.  Explain  why  this  can  not  be 
done. 

149.  Theorem  XVIII.     A  tangent  is  perpendicular  to  the 
radius  drawn  to  the  point  of  tangency. 

150.  Theorem  XIX.     A  line  perpendicular  to  a  radius  at  its 
extremity  on  the  circumference  is  tangent  to  the  circle. 

151.  Theorem  XX.     The  center  of  a  circle  lies  on  a  perpen- 
dicular to  a  tangent  at  the  point  of  tangency. 


iv,  §  152]  INSCRIBED  ANGLES  129 

Suggestion.  Suppose  that  the  center  does  not  lie  on  the 
perpendicular.  From  the  center  draw  a  radius  to  the  point 
of  tangency.  Prove  that  this  radius  coincides  with  the  per- 
pendicular, and  hence  that  the  center  lies  on  the  perpen- 
dicular. 

152.  Theorem  XXI.     The  perpendicular  from  the  center  of 
a  circle  to  a  tangent  meets  it  at  the  point  of  tangency. 

Suggestion.  Suppose  that  the  perpendicular  does  not  go 
through  the  point  of  tangency.  Draw  a  radius  to  the  point 
of  tangency  and  prove  that  it  coincides  with  the  perpen- 
dicular, and  hence  that  the  perpendicular  goes  through  the 
point  of  tangency. 

153.  Exercises. 

1.  If  the  distance  from  a  point  to  the  center  of  a  circle 
equals  the  diameter,  determine  the  size  of  the  angle  between 
the  tangents  drawn  from  the  point  to  the  circle.     Prove. 
Apply  Art.  72,  Ex.  5. 

2.  If  two  tangents  are  drawn  from  a  point  to  a  circum- 
ference of  a  circle  the  angle  contained  by  them  is  supple- 
mentary to  the  angle  contained  by  the  radii  drawn  to  the 
point  of  tangency. 

3.  If  two   diameters  are  drawn  at  right  angles  to  one 
another,  and  tangents  are  drawn  through  their  ends,  the  figure 
formed  by  the  tangents  is  a  square. 

4.  If  three  diameters  of  a  circle  are  drawn  so  that  they 
divide  the  perigon  about  the  center  into  six  equal  parts,  and 
if  tangents  are  drawn  through  their  ends,  a  regular  hexagon 
will  be  formed. 

5.  If  in  Exercise  4  you  increase  the  number  of  diameters 
and  tangents,  variable  quantities  will  appear. 

What  is  the  independent  variable? 
What  are  the  dependent  variables? 
What  is  the  constant? 
Give  answer  using  the  word  function. 


130  PLANE  GEOMETRY  [IV,  §  154 

Examine  the  variables  and  tell  whether  the  dependent 
variables  increase  or  decrease  as  you  change  your  inde- 
pendent variable. 

If  they  increase  tell  how  large  they  may  become.  If  they 
decrease  tell  how  small  they  may  become. 

6.  If  two  tangents  to  a  circle  are  parallel,  the  line  join- 
ing their  points  of  contact  is  a  diameter. 

7.  A  circle  tangent  to  the  three  sides  of  a  square  must  be 
tangent  to  the  fourth  side. 

8.  Two  circles  of  radii  1  and  2  inches  respectively  have 
the  same  center.     A  chord  of  the  larger  circle  is  tangent  to 
the  smaller.     How  long  is  the  chord? 

154.  Definition.     A  segment  of  a  circle  is  either  of  the  two 
portions  into  which  a  chord  divides  the 
circle. 

The  portion  Si  is  called  the  minor  seg- 
ment. 

The  portion  S  is  called  the  major  seg- 
ment. 

If  two  segments  are  equal  they  are  semicircles. 

An  angle  formed  by  two  chords  meeting  on  a  circumference 
is  called  an  inscribed  angle. 

The  arc  which  the  arms  of  an  inscribed  angle  cut  off  is  said 
to  subtend  the  angle,  and  the  angle  is  said  to  stand  on  or 
intercept  that  arc. 

In  the  figure,  angle  ABC  is  said  to  stand 
on,  or  intercept,  arc  AC  and  arc  AC  is  said 
to  subtend  the  inscribed  angle  ABC. 

If  a  chord  is  drawn  from  A  to  C,  we  say 
that  angle  ABC  is  inscribed  in  the  major 
segment  AC  and  intercepts  the  minor 
arc  AC. 


IV,  §  155) 


INSCRIBED  ANGLES 


131 


155.  Theorem  XXII.     An  inscribed  angle  equals  half  the 
central  angle  standing  on  the  same  arc. 


Given  the  angle  ABC  inscribed  in  the  circle  whose  center 
is  0,  and  the  central  angle  AOC,  standing  on  the  same  arc  AC. 

To  prove  that  angle  ABC  is  equal  to  half  angle  AOC. 

Analysis.  There  are  three  possibilities.  One  arm  of  the 
angle  may  be  a  diameter,  or  both  arms  may  lie  on  one  side  of 
the  diameter,  or  one  arm  may  be  on  one  side  of  the  diameter 
and  the  other  on  the  other  side,  as  shown  in  the  three  figures 
above. 

In  the  first  figure  we  can  prove  that  Z  AOC  is  twice  Z  ABC, 
if  we  can  prove  that  it  is  equal  to  the  sum  of  A  ABC  and 
BCO,  and  can  then  show  that  A  ABC  and  BCO  are  equal. 
We  can  show  that  A  ABC  and  BCO  are  equal  if  we  can  show 
that  A  OCB  is  isosceles. 

Proof.    (First  figure.)     A  OCB  is  isosceles.  Why? 

Z  ABC  =  Z  BCO.  Why? 

Z  AOC  is  an  exterior  angle  to  A  OCB. 

Z  AOC  =  Z  OBC  +  Z  BCO.  Why? 

Z  AOC  =2  Z  ABC.  Why? 

Z  ABC  =  J  Z  AOC. 
State  the  theorem. 

In  the  second  figure,  to  prove  Z  AOC  =  2  Z  ABC,  we 
make  application  of  the  truth  proved  in  the  first  figure.  Draw 
in  diameter  BR,  and  show  that  Z  ABC  is  the  result  of  sub- 


132  PLANE  GEOMETRY  [IV,  §  156 

tracting  Z  RBA  from  Z  RBC,  and  Z  AOC  is  the  result  of 
subtracting  Z  ROA  from  Z  .ROC. 

Give  this  proof. 

In  the  third  figure,  after  drawing  in  the  diameter,  show  that 
Z  ABC  =  Z  ABR  +  Z  #BC,  and  Z  AOC  =  Z 
Z  ROC. 

Give  this  proof. 

Corollary.  Angles  inscribed  in  the  same  segment 
or  in  equal  segments  are  equal.  Give  reason. 

Exercise.  If  chord  BC  is  taken  shorter  and  shorter,  what 
change  will  take  place  in  the  angle  inscribed  in  the  minor 
segment?  In  the  angle  inscribed  in  the  major  segment? 

156.  Theorem  XXIII.     An  angle  inscribed  in  a  segment  is 
greater  than,  equal  to,  or  less  than  a  right  angle,  according  as  the 
segment  is  less  than,  equal  to,  or  greater  than  a  semicircle. 
Supply  proof. 

157.  Definition.    A  polygon  is  said  to  be  inscribed  in  a 
circle  when  its  sides  are  chords  of  the  circle. 

We  also  say  that  the  circle  is  circumscribed 
about  the  polygon. 

A  polygon  is  said  to  be  circumscribed  about 
a  circle,  when  its  sides  are  tangent  to  the  circle. 
We  also  say  that  the  circle  is  inscribed  in  the  polygon. 

158.  Exercises. 

1.  If  an  inscribed  angle  is  20  degrees,  how  many  degrees  in 
the  intercepted  arc?     If  it  is  50  degrees?     If  110  degrees?    If 
178  degrees? 

2.  Angle  ABC  is  inscribed  in  a  circle.     If  arc  AC  is  the 
length  of  a  radius,  what  is  the  value  of  angle  ABC? 

3.  What  arc  is  intercepted  by  an  inscribed  angle  which 
equals  a  radian? 

4.  Draw  an  inscribed  angle  which  will  equal  J  TT  radians. 


IV,  §  158]  INSCRIBED  ANGLES  133 

5.  In  the  figure  in  Exercise  2  let  the  points  B  and  A  remain 
fixed  and  let  the  point  C  move  around  the  circumference  from 
A  to  B.     State  the  least  value  of  Z  ABC.    How  large  may 
it  become? 

6.  If  points  A  and  C  remain  fixed  and  point  B  moves  from 
AtoC  tell  all  the  values  that  angle  ABC  may  have. 

7.  If  two  equal  arcs  are  laid  off  on  a  circumference,  and 
their  ends  are  joined,  there  will  be  two  chords  which  intersect 
and  two  chords  which  are  parallel.    Prove  that  the  chords  that 
intersect  are  equal  and  divide  each  other  into  mutually  equal 
parts,  and  also  prove  that  the  other  two  chords  are  parallel. 

8.  A  triangle,  one  of  whose  angles  is  25  degrees,  and  another 
angle  100  degrees,  is  inscribed  in  a  circle.     Tell,  with  reason, 
the  number  of  degrees  in  the  arcs  subtended  by  the  sides  of  the 
triangle. 

9.  With  the  extremities  of  any  diameter  as  centers  and  with 
a  radius  equal  to  the  radius  of  the  circle  arcs  are  struck  cutting 
the  circumference.     Show  that  the  points  so  obtained  to- 
gether with  the  extremities  of  the  diameter  form  the  vertices 
of  a  regular  inscribed  hexagon,  by  showing  that  the  angles 
are  equal  and  the  sides  are  equal. 

Calculate  the  perpendicular  distance  from  the  center  of  the 
circle  to  one  of  the  sides,  if  radius  of  the  circle  is  r. 

10.  Tell  the  number  of  degrees  in  each  of  the  arcs  sub- 
tended by  the  side  of  a  regular  hexagon  inscribed  in  a  circle; 
of  a  nonagon;  of  a  pentagon. 

11.  Prove  that  the  sum  of  the  angles  of  a  triangle  is  equal 
to  a  straight  angle  by  inscribing  it  in  a  circle. 

12.  Prove  similarly  that  the  opposite  angles  of  an  inscribed 
quadrilateral  are  supplementary. 

13.  If  a  circle  is  drawn  with  one  side  of  an  isosceles  triangle 
as  a  diameter,  it  will  bisect  the  base. 

14.  Two  chords  perpendicular  to  a  third  chord  at  its  ex- 
tremities are  equal. 


134  PLANE  GEOMETRY  [IV,  §  159 

15.  If  any  number  of  angles  are  inscribed  in  the  same  seg- 
ment of  a  circle,  their  bisectors  meet  in  a  point  on  the  sub- 
tending arc. 

We  may  use  the  fact  that  an  angle  inscribed  in  a  semicircle 
is  a  right  angle,  to  construct  a  tangent  to  a  circle  at  a  given 
point  on  its  circumference. 

159.  Problem  I.  To  construct  a  tangent  to  a  circle  at  a  given 
point  on  its  circumference. 


Given  the  circle  whose  center  is  Q,  with  the  given  point  A 
on  the  circumference. 

To  construct  a  tangent  at  point  A. 

Construction.  Draw  OA.  With  any  center  M  (not  on 
OA)  and  a  radius  MA  draw  a  circle.  Call  the  point  where  it 
cuts  OA,  R.  Draw  diameter  RM,  cutting  the  circle  again  at 
P.  AP  is  tangent  at  A. 

Proof.     Give  the  proof. 

Also  solve  this  problem  by  using  Art.  47,  Cor.  1. 

Exercises. 

1.  Show  how  to  circumscribe  an  equilateral  triangle  about 
a  given  circle. 

2.  Show  how  to  circumscribe  a  square  about  a  circle 

3.  Divide  a  circumference  into  6  equal  parts  by  drawing 
chords  equal  to  the  radius. 

Draw  tangents  at  the  points  of  division.  What  kind  of  a, 
polygon  is  formed  by  these  tangents? 


IV,  §  160]  INSCRIBED  ANGLES  135 

160.  Problem  II.     To  draw  a  tangent  to  a  circle  from  an  ex- 
ternal point. 


Given  the  circle  whose  center  is  0,  and  the  external 
point  A. 

To  draw  a  tangent  to  the  circle  passing  through  point  A. 

Analysis.  Since  the  tangent  from  A  must  meet  the  radius 
from  0  at  right  angles,  and  since  to  get  this  right  angle,  we 
may  construct  a  semicircle  which  will  contain  it,  namely  a 
semicircle  with  OA  as  a  diameter,  bisect  OA  and  with  the 
mid-point  as  a  center  and  one  half  OA  as  a  radius  draw  a 
circle  cutting  the  given  circle  in  T  and  TI.  Draw  A  T  and 
AT,. 

Proof.     Give  the  pfoof. 

Exercises. 

1.  Show  that  the  tangents  from  an  exterior  point  to  a  cir- 
cle are  equal. 

2.  Draw  a  circle  with  radius  2  inches  long.     Mark  an  ex- 
terior point  1  inch  from  the  circumference.     Construct  the 
tangents  from  this  point  to  the  circle.     Calculate  the  lengths 
of  these  tangents.     Check  by  measurement  of  drawing. 

3.  If  the  radius  of  a  circle  is  3  cm.  long,  how  far  from  the 
center  must  a  point  be  taken  to  make  the  tangents  from  the 
point  12  cm.  long? 

4.  If  two  exterior  points  are  unequally  distant  from  the 
center,  the  tangents  from  the  more  remote  point  are  the 
longer. 


136 


PLANE  GEOMETRY 


(IV,  §  161 


161.  Theorem  XXIV.  An  angle  formed  by  a  tangent  and  a 
chord  is  equal  to  one-half  the  central  angle  standing  on  the  same 
arc. 


Given  the  circle  with  center  0,  with  the  angle  XMR  formed 
by  the  tangent  YX  and  the  chord  MR  meeting  at  the  point  M. 
Also  given  the  central  angle  MOR  standing  on  the  inter- 
cepted arc  MR. 

To  prove  that  angle  XMR  is  equal  to  one  half  of  angle  MOR. 

Analysis.  In  order  to  prove  that  angle  XMR  is  one-half 
of  angle  MOR,  extend  the  radius  MO  through  0  making 
diameter  MK. 

Since  Z  XMR  is  the  difference  between  right  angle  XMK 
and  inscribed  angle  RMK,  we  have  but  to  show  that  Z  MOR 
is  equal  to  the  difference  between  a  straight  angle  of  which 
right  angle  XMK  must  be  one-half,  and  Z  ROK  of  which  in- 
scribed angle  RMK  is  one-half. 

Proof. 

Str.  Z  MOK  -  Z  ROK  =  Z  MOR.  Why? 

Z  XMK  =  \  Z  MOK.  Why? 

Z  RMK  =  \  Z  ROK.  Why? 

/.    /  XMK  -  Z  RMK  =  I  (Z  MOK  -  Z  ROK).  Why? 

Z  XMR  =  \  Z  MOR.  Why? 

State  the  proposition  proved. 


IV,  §  162]  INSCRIBED  ANGLES  137 

Corollary.     Tangents  to  a  circle  from  an  exterior  point  are 
equal.     Why? 
Exercises. 

1.  In  the  figure  of  Theorem  XXIV,  Z  RM  Y  is  also  formed 
by  a  tangent  and  a  chord.     Prove  that  Z  RM  Y  equals  half  of 
the  reflex  angle  .ROM,  which  subtends  the  major  arc  RM. 

2.  In  the  figure  of  Theorem  XXIV  suppose  R  to  move 
around  on  the  circumference.    What  angles  will  change  value? 

162.  Theorem  XXV.     An  angle  formed  by  two  chords  in- 
tersecting within  a  circle,  is  equal  to  one-half  of  the  sum  of  the 
central  angles  subtended  by  the  intercepted 

arcs. 

Suggestion.  Draw  a  chord  from  one 
end  of  one  of  the  chords  to  one  end  of 
the  other.  Show  that  the  angle  formed 
by  the  chords  is  equal  to  the  sum  of  the 
two  inscribed  angles  standing  on  the  in- 
tercepted arcs  (Art.  65,  Cor.  1)  and  hence  equal  to  half  tLe 
sum  of  the  central  angles  on  these  arcs. 

163.  Theorem  XXVI.     An  angle  formed  by  two  secants  in- 
tersecting without  the  circle,  is  equal  to  one-half  the  difference 
between  the  central  angles  standing  on  the  intercepted  arcs. 

Suggestion.     Draw  a  chord  from  the  point  where  one  of  the 
secants  cuts  the  circumference 
to  the  point  where  the  other 
secant  cuts  the  circumference. 

Show  that  the  angle  formed 
by  the  secants  is  equal  to  the 
difference  between  the  two  in- 
scribed angles  standing  on  the 
intercepted  arcs,  Art.  65  Cor. 
1 ,  and  hence  equal  to  half  the 
difference  between  the  central  angles  standing  on  the  inter- 
cepted arcs. 


138 


PLANE  GEOMETRY 


[IV,  §  164 


164.  Theorem  XXVII.     An  angle  formed  by  a  secant  and  a 
tangent  is  equal  to  one-half  the  difference  between  the  central 
angles  standing  on  the  intercepted  arcs. 

Suggestion.  Show  that  by  turning  one  of  the  secants  in 
the  figure  in  Theorem  XXVI,  about  the  point  of  intersection, 
until  points  C  and  B  come  together,  that  theorem  still  holds 
for  every  position  of  the  secant  until  points  C  and  B  coincide, 
and  hence  holds  when  they  coincide. 

This  suggests  that,  by  regarding  the  tangent  as  a  special 
position  of  a  secant,  we  may  say  that  a  tangent  cuts  a  circle 
in  two  coincident  points. 

Exercises. 

1.  The  angle  between  two  chords  AB  and  CD  is  40°;  arc 
AD  is  50°.     How  many  degrees  in  arc  BCf    What  is  the 
angle  between  secants  BD  and  CA  ? 

2.  The  angle  between  two  secants  is  20° ;  one  of  the  inter- 
cepted arcs  is  100°;  how  many  degrees  in  the  other? 

165.  Theorem  XXVIII.     //  two  parallel  lines  intercept  arcs 
on  a  circumference,  the  intercepted  arcs  are  equal. 

These  parallel  lines  may  have  three  different  positions. 


A.B, 


A.B 


C,D 


Given  parallel  lines  PI  and  P2,  cutting  the  circle  whose 
center  is  0  in  points  A,  B  and  C,  D  respectively,  intercepting 
arcs  CA  and  BD. 

To  prove  that  arcs  CA  and  BD  are  equal. 

Analysis.  To  prove  that  arcs  CA  and  BD  are  equal,  draw 
diameter  H  K  perpendicular  to  PI,  and  by  showing  that  HK 


IV,  §  166] 


INSCRIBED  ANGLES 


139 


is  also  perpendicular  to  P2,  show  that  arc  C  K  equals  arc  KD, 
and  arc  A  K  equals  arc  KB. 

Give  the  proof.     Explain  how  the  same  proof  answers  for 
all  three  figures. 

166.  Theorem  XXIX.     //  two  circumferences  intersect,  their 
line  of  centers  is  the  perpendicular  bisector  of  their  common  chord. 


Suggestion.  Suppose  that  the  line  of  centers  is  not  per- 
pendicular to  chord.  Draw  radii  from  each  center  per- 
pendicular to  the  chord.  Prove  that  line  of  centers  coincides 
with  these. 

By  thinking  of  these  circles  gradually  moving  apart,  state 
and  prove  a  proposition  about  line  of  centers  and  common 
tangent. 

167.  In  the  exercises  below,  the  following  definitions  will 
be  needed. 

Definitions.  Concentric  circles  are  circles  with  a  common 
center. 

Two  circles  are  tangent  to  each  other  when  they  have  but 
one  point  in  common;  internally  tangent  when  they  lie  on  the 
same  side  of  the  tangent  at  their  common  point;  externally 
tangent  when  they  lie  on  opposite  sides  of  that  tangent  line. 


Concentric  Circles. 


Internally  Tangent. 


Externally  Tangent. 


140 


PLANE  GEOMETRY 


[[IV,  §  168 


168.  Exercises. 

1.  Two  secants  are  drawn  from  point  P,  the  first  cutting 
the  circumference  in  T  and  R,  the  second  in  Q  and  S.     If 
arc  TQ  is  50°  and  arc  RS  is  150°,  how  many  degrees  are 
there  in  the  angle  between  the  secants?     Solve  again  if  arc 
TQ  is  half  a  radian  and  arc  RS  three  times  as  large. 

2.  Let  ABC  be  an  inscribed  angle  with  one  arm  BC  a 
diameter  whose  center  is  0.     OR  is  drawn  so  that  arc  AR 
equals  arc  RC.     Prove  that  line  OR  is 

parallel  to  BA. 

3.  Prove  Theorem  XXV  by  the  adjoin- 
ing figure.     Use  Articles  59  and  165. 

Line  QD  is  drawn  parallel  to  line  AB. 

4.  Prove  Theorem  XXVI  by  the  adjoin- 
ing figure. 

Line  PQ  is  drawn  parallel  to  line  AB. 

5.  If  two  tangents  are  drawn  to  a  circle 
from  the  same  point,  prove  that  they  are 
equal  by  drawing  in  the  chord  of  tan- 

gency,  and  proving  that  two  of  the  angles  formed  are  equal, 
and  hence  that  the  triangle  formed  is  isosceles. 

6.  If  the  angle  formed  by  two  tangents  from  an  external 
point  to  a  circle  is  10°,  find  the  size  of  the  arc  subtended  by 
the  chord  of  tangency. 

7.  By  inscribing   an   isosceles   triangle  in  a  circle,  and 
drawing  tangents  at  its  vertices,  prove  a  new  isosceles  tri- 
angle is  formed. 

8.  An  angle  formed  by  two  tangents  drawn  from  an  ex- 
terior point  to  a  circle,  is  equal  to  one-half  the  difference 
between  the  central  angles  standing  on  the  intercepted  arcs. 
Show  this  to  be  a  special  case  of  Theorem  XXVI. 

9.  An  angle  formed  by  two  tangents  from  an  exterior 
point  to  a  circle  is  equal  to  the  supplement  of  the  angle 
formed  by  the  radii  to  the  points  of  contact. 


IV,  §168]  INSCRIBED  ANGLES  141 

10.  Can  a  circle  be  circumscribed  about  a  rhombus? 

11.  If   two   circles  are  tangent,   and  two   perpendicular 
secants  are  drawn  through  the  point  of  tangency,  lines  join- 
ing the  ends  of  the  secants  are  diameters. 

12.  If  three  angles  are  drawn  standing  on 
the  same  chord,  one  with  its  vertex  outside 
of  the  circumference,  another  with  its  ver- 
tex on  the   circumference,  and  the  third 
with  its  vertex  within  the  circle,  the  one 

with  its  vertex  outside  will  be  less  than,  and  the  one  with  its 
vertex  inside  will  be  greater  than,  the  inscribed  angle. 

13.  Two   straight   pieces  of    track,  which   — __^ 
make  equal  angles  with  the  line  joining  their 

end  points,  are  to  be  joined  by  a  circular  arc. 

To  avoid  jar  when  a  car  rounds  the  turn  the 

parts  of  the  track  must  be  tangent  to  the  arc.     Show  how 

to  construct  such  an  arc. 

14.  If  two  circles  are  concentric,  all  chords  of  the  one, 
tangent  to  the  other,  are  equal.     Hence  show  that  a  quadri- 
lateral which  is  circumscribed  about  the  one  circle  and  in- 
scribed in  the  other  must  be  a  square. 

15.  If  two  circles  are  tangent  externally 
and  a  secant  is  drawn  through  the  point 
of  tangency,  the  tangents  drawn  to  the 
circles  at   the   ends   of   the   secant  are 

parallel.  Show  that  the  two  minor  arcs,  or  the  two  major 
arcs,  subtend  equal  angles  at  the  centers  of  their  respective 
circles. 

16.  If  two  circles  are  tangent  and 
two  secants  are  drawn  through  the 
point  of  tangency,  the  lines  joining 
the  ends  of  the  secants  are  parallel. 

17.  If  two  circles  are  tangent  and  three  secants  are  drawn 
through  the  point  of  tangency,  the  lines  joining  their  ends 
will  form  two  mutually  equiangular  triangles. 


142 


PLANE  GEOMETRY 


[IV,  §  168 


18.  An  inscribed  trapezoid  is  isosceles.    Prove  this.    Also 
apply  Theorems  XXII  and  XXVIII  to  show  that  the  angles 
at  the  extremities  of  either  base  are  equal. 

19.  If  two  circles  intersect  and  secants  are  drawn  through 
the  points  of  intersection,  the  lines  joining  the  ends  of  the 
secants  are  parallel. 

Suggestion.  Draw  the  common  chord  and 
prove  that  the  angle  A  and  angle  C  are  sup- 
plementary by  proving  that  they  are  respect- 
ively supplementary  to  angles  at  E. 

20.  Prove  that  the  sum  of  the  angles  of  a  triangle  is  equal 
to  a  straight  angle  by  inscribing  the  triangle  in  a  circle  and 
drawing  a  tangent  to  the  circles  at  one  of  the  vertices. 

21.  If  a  hexagon  is  circumscribed  about  a  circle,  the  sum 
of  the  first,  third  and  fifth  sides  is  equal  to  the  sum  of  the 
second,  fourth  and  sixth  sides. 

22.  Is  it  true  for  any  circumscribed  polygon  that  the  sum 
of  one  alternate  set  of  sides  is  equal  to  the  sum  of  the  other 
alternate  set  of  sides? 

23.  By  applying  Theorem  XXII,  show  that  if  an  inscribed 
polygon  is  equi-lateral,  it  is  also  equi-angular.     Also  prove 
the  converse.     Prove  the  same  for  a  circumscribed  polygon. 

24.  Construct  a  Gothic  arch  and  con- 
struct tangent  lines  from  a  point  directly 
over  the  center  of  the  arch. 

If  the  arch  is  12  feet  wide,  and  the 
tangents  are  drawn  from  a  point  20  feet 
above  the  middle  of  the  base,  find  the 
length  of  each  tangent  line. 

Suggestion.  To  construct  tangent  line 
DK,  draw  a  circle  on  BD  as  diameter  and 
let  this  circle  cut  arc  AH  at  K.  Then 
DK  will  be  tangent  to  arc  AH  because 
90°. 


BKD 


To  calculate  DK,  use 


rt.  A  BDK,  in  which  you  can  easily  find  BK  and  BD. 


IV,  §  168] 


INSCRIBED  ANGLES 


143 


25.  In  the  following  figure,  the  circles  Oi,  02,  03  are  all 
tangent  to  the  line  XY  at  the  point  T.  AK  and  BC  are 
chords  of  circle  whose  center  is  Oi  and  tangent  to  the  circle 
whose  center  is  02  at  the  points  E  and  F  respectively.  Secants 
TE,  TG,  TF  are  drawn,  cutting  the  circumference  whose 
center  is  Oi  in  the  points  P,  B,  H  respectively,  and  the  cir- 
cumference whose  center  is  Oa  in  the  points  M,  D,  Q,  respect- 
ively.  Lines  are  drawn  joining  these  points.  Also  lines  EG, 
GF,  EF}  AB,  AC,  are  drawn. 

Name  all  the  pairs  of  equal  arcs  that  you  can  find  in  the 
figure,  and  tell  reason  for  knowing  them  to  be  equal.  Name 
all  the  pairs  of  equal  angles  that  you  can  find,  telling  reason 
for  knowing  them  equal.  A  pair  of  equal  arcs  that  is  apt 
to  be  overlooked  is  BH  and  HC. 


144  PLANE  GEOMETRY  [IV,  §  169 

169.  Summary  of  Chapter  IV.  , 

Part  I.     Central  Angles,  Arcs,  Chords. 

Definitions.     Circle,  radius,  diameter,  chord,  equal  circles,  arc,  semi- 
circle, semicircumference,  sector,  central  angle. 
Postulates. 

1.  A  circumference  is  determined  when  its  center  and  radius  are 
known. 

2.  A  point  is  within,  on,  or  without  a  circle  according  as  its  distance 
from  the  center  is  less  than  equal  to  or  more  than  the  radius. 

Propositions  on  central  angles  and  arcs. 

Theorem  I.  In  the  same  or  equal  circles  if  two  central  angles  are 
equal,  the  arcs  on  which  they  stand  are  equal. 

Corollary  1.  A  diameter  divides  a  circumference  into  two  equal 
parts. 

Corollary  2.  In  the  same  or  equal  circles  sectors  which  have  equal 
angles  are  equal. 

Corollary  3.     A  diameter  divides  a  circle  into  equal  parts. 

Theorem  II.  In  the  same  or  equal  circles  if  two  central  angles  stand 
on  equal  arcs,  the  angles  are  equal. 

Definition  of  measurement  of  arcs. 

Theorem  III.  In  the  same  or  equal  circles  if  two  central  angles  are 
unequal,  the  arcs  on  which  they  stand  are  unequal  and  the  greater 
angle  stands  on  the  greater  arc. 

Theorem  IV.  In  the  same  or  equal  circles  if  central  angles  stand  on 
unequal  arcs,  the  angles  are  unequal,  the  greater  angle  standing  on  the 
greater  arc. 

Functions.     Definition  and  Illustrations. 

Theorems  on  Arcs  and  Chords. 

Theorem  V.  In  the  same  or  equal  circles  if  two  arcs  are  equal  they 
are  subtended  by  equal  chords. 

Theorem  VI.  In  the  same  or  equal  circles  if  two  chords  are  equal 
they  subtend  equal  arcs. 

Theorem  VII.  In  the  same  or  equal  circles  if  two  minor  arcs  are 
unequal  they  are  subtended  by  unequal  chords,  and  the  greater  arc  is 
subtended  by  the  greater  chord. 

Theorem  VIII.  In  the  same  or  equal  circles  if  two  chords  are  un- 
equal, they  subtend  unequal  minor  arcs,  and  the  less  chord  subtends 
the  less  minor  arc. 


IV,  §  169]  INSCRIBED  ANGLES  145 

Summary  of  the  above  illustrating  functions,  independent  and  de- 
pendent variables. 

Theorem  IX.  A  diameter  which  is  perpendicular  to  a  chord  bisects 
the  chord  and  its  subtended  arc. 

Theorem  X.     A  diameter  which  bisects  a  chord  is  perpendicular  to  it. 

Theorem  XI.  The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle  and  bisects  the  subtended  arc. 

Corollary.     In  general  two  chords  cannot  bisect  each  other. 

Theorems  on  Chords  and  Distances  from  the  Center. 

Theorem  XII.  In  the  same  or  equal  circles  equal  chords  are  equi- 
distant from  the  center. 

Theorem  XIII.  In  the  same  or  equal  circles  if  two  chords  are  equi- 
distant from  the  center,  they  are  equal. 

Theorem  XIV.  In  the  same  or  equal  circles  if  two  chords  are  un- 
equal, they  are  unequally  distant  from  the  center;  the  greater  chord  is 
the  less  distant. 

Theorem  XV.  In  the  same  or  equal  circles  if  two  chords  are  un- 
equally distant  from  the  center,  they  are  unequal;  the  chord  less  distant 
is  the  greater. 

Corollary.     The  diameter  is  the  longest  chord  in  a  circle. 

Summary  of  the  above,  illustrating  functions,  independent  and  de- 
pendent variables. 
Propositions  on  Tangents. 

Theorem  XVI.  Of  all  lines  drawn  through  a  point  on  the  circum- 
ference, the  perpendicular  to  the  radius  drawn  to  that  point  is  the  only 
one  that  does  not  meet  the  circumference  again. 

Theorem  XVII.  One  and  only  one  tangent  can  be  drawn  to  a  circle 
at  a  point  on  its  circumference. 

Theorem  XVIII.  A  tangent  is  perpendicular  to  a  radius  drawn  to 
the  point  of  tangency. 

Theorem  XIX.  A  line  perpendicular  to  a  radius  at  its  extremity  on 
the  circumference  is  tangent  to  the  circle. 

Theorem  XX.  The  center  of  a  circle  lies  on  the  perpendicular  to  a 
tangent  at  the  point  of  tangency. 

Theorem  XXI.  The  perpendicular  from  the  center  of  a  circle  to  a 
tangent  meets  it  at  the  point  of  tangency. 

Part  II.     Inscribed  Angles. 

Definitions.     Segments  of  a  circle.     Inscribed  angle. 

Theorem  XXII.  An  inscribed  angle  is  equal  to  half  the  central  angle 
standing  on  the  same  arc, 


146  PLANE  GEOMETRY  [iv,  §  169 

Corollary.     Angles  inscribed  in  the  same  or  equal  segments  are  equal. 

Theorem  XXIII.  An  angle  inscribed  in  a  segment  is  greater  than, 
equal  to  or  less  than  a  right  angle,  according  as  the  segment  is  less  than, 
equal  to,  or  greater  than  a  semicircle. 

Definition  of  an  inscribed  polygon. 

Problem  I.  To  construct  a  tangent  to  a  circle  at  a  given  point  on  its 
circumference. 

Problem  II.  To  construct  a  tangent  to  a  circle  that  shall  pass  through 
a  given  external  point. 

Theorem  XXIV.  An  angle  formed  by  a  tangent  and  a  chord  is  equal 
to  half  the  central  angle  standing  on  the  intercepted  arc. 

Corollary.     Tangents  to  a  circle  from  the  same  point  are  equal. 

Theorem  XXV.  An  angle  formed  by  two  chords  intersecting  within 
a  circle,  is  equal  to  half  the  sum  of  the  central  angles  standing  on  the 
intercepted  arcs. 

Theorem  XXVI.  An  angle  formed  by  two  secants  intersecting  with- 
out the  circle  is  equal  to  half  the  difference  between  the  central  angles 
standing  on  the  intercepted  arcs. 

Theorem  XXVII.  An  angle  formed  by  a  secant  and  a  tangent  is 
equal  to  half  the  difference  between  the  central  angles  standing  on  the 
intercepted  arcs. 

Theorem  XXVIII.  If  two  parallel  lines  intercept  arcs  on  a  circum- 
ference, the  arcs  are  equal. 

Theorem  XXIX.  If  two  circumferences  intersect,  their  line  of  cen- 
ters is  the  perpendicular  bisector  of  their  common  chord. 

1.  Into  what  two  parts  is  the  subject  matter  of  this  chapter  divided? 

2.  Is  the  subject  matter  of  the  different  parts  closely  related? 

3.  Examine  the  theorems  of  this  chapter.     Select  some  that  have 
been  proved  without  one  or  more  of  the  preceding  chapters?     Explain 
your  selection  by  giving  an  outline  of  the  proof. 

4.  Are  there  any  theorems  of  the  second  part  that  have  no  dependence 
on  those  ot  the  first  part? 

6.  Of  the  solids  which  you  examined  in  Chapter  I,  which  do  you  think 
might  be  so  cut  by  a  plane  that  the  section  would  be  circular? 

6.  State  some  of  the  exercises  that  you  consider  to  have  some  con- 
nection with  real  things  around  you. 

The  theorems  of  this  chapter  are  selected  from  Book  III  of  Euclid's 
Geometry. 


CHAPTER  V 

PART  I— LOCI  OF  POINTS.  PART  II— COORDINATE 
GEOMETRY. 

PART  I— LOCI  OF  POINTS 

170.  General  Discussion.  In  order  to  give  the  location 
of  a  place  it  is  necessary  to  tell  its  position  with  reference  to 
some  other  places  whose  locations  are  known  to  both  the 
person  seeking  the  information  and  the  person  giving  it. 
In  fact  it  will  be  found  that  in  order  to  state  definitely  the 
location  of  any  place,  three  distinct  items  of  information 
concerning  that  place  must  be  mentioned. 

For  example,  I  have  in  mind  a  certain  point  in  this  room. 
Can  you  tell  me  the  exact  location  of  the  point?  It  is  two 
feet  from  the  north  wall.  Can  you  now  determine  it  more 
definitely  than  you  could  before?  I  say  that  it  is  four  feet 
from  the  ceiling.  Can  you  now  tell  the  exact  location  of  the 
point?  I  give  the  further  information  that  it  is  equidistant 
from  the  east  and  west  walls.  What  is  the  exact  location  of 
the  point? 

Again,  suppose  that  you  wish  to  tell  the  location  of  the 
town  in  which  you  live.  You  state  that  its  latitude  is  so 
many  degrees  and  that  its  longitude  is  so  many  degrees,  say, 
41  degrees  north  latitude  and  73  degrees  west  longitude. 
This  description  involves  the  knowledge  of  an  imaginary 
line  called  the  equator  from  which  to  count  the  latitude  41 
degrees  north,  and  the  knowledge  of  an  imaginary  line  north 
and  south  through  the  town  of  Greenwich,  from  which  to 
count  the  longitude  73  degrees  west.  There  is  the  further 

147 


148  PLANE  GEOMETRY  [V,  §  171 

information  that  the  town  is  on  the  earth's  surface,  which  is  a 
third  fact,  telling  us  the  distance  that  the  town  is  from  the 
center  of  the  earth. 

To  locate  the  position  of  an  observatory  situated  at  the  top 
of  a  mountain,  we  would  give  its  latitude,  its  longitude,  and 
its  height  above  sea  level. 

171.  Locus.     It  is  seen  readily  that  when  but  one  fact  is 
mentioned  in  our  description,  there  exist  a  great  many  points 
that  answer  the  description.     In  the  illustration  of  the  loca- 
tion of  a  point  in  the  room,  there  is  a  whole  plane  of  points 
that  answer  the  description  "two  feet  from  the  north  wall." 
After  we  added  to  our  description  "four  feet  from  the  ceiling/' 
there  was  still  left  a  whole  line  of  points  answering  the  given 
descriptions. 

Definition.  The  line  or  group  of  lines,  the  plane  or  group 
of  planes,  every  point  of  which  satisfies  certain  given  con- 
ditions, and  on  which  every  point  that  satisfies  these  con- 
ditions lies,  is  called  the  locus  of  the  point  that  satisfies  these 
conditions. 

Whenever  we  wish  to  prove  that  a  line  is  the  locus  of  a 
point  satisfying  a  given  condition,  we  must  prove  two  dis- 
tinct things. 

First,  every  point  on  the  line  satisfies  the  given  condition. 
Second,  every  point  satisfying  the  given  condition  is  on  the 
line.  The  one  of  these  is  just  as  important  as  the  other. 

We  shall  now  take  up  some  theorems  involving  a  few  simple 
loci. 

172.  Theorem  I.     The  locus  of  a  point  at  a  given  distance 
from  a  given  fixed  point  is  a  circumference  about  the  given  fixed 
point  as  a  center,  with  the  given  distance  as  a  radius. 

This  follows  immediately  from  the  definition  of  circum- 
ference, Art.  122,  and  Postulate  VIII,  Art.  123,  and  might 
have  been  given  as  a  corollary  to  the  statements  there 
made. 


v,  §  1731  LOCI  149 

173.  Theorem  II.  The  locus  of  a  point  equidistant  from 
two  fixed  points  is  the  perpendicular  bisector  of  the  line  that 
joins  them. 

Y 

P 


Kx 

Given  the  two  fixed  points  M  and  N.  Also  line  YYl  the 
perpendicular  bisector  of  line  MN. 

To  prove  that  line  YYi  is  the  locus  of  a  point  equidistant  from 

M,N. 

Analysis.  In  order  to  prove  this  we  shall 'prove  that  P, 
any  point  on  YYi,  is  equidistant  from  M  and  N,  and  that 
PI,  any  point  not  on  Y  FI,  is  not  equidistant  from  M  and  N. 
In  other  words  we  shall  prove  that  PM  equals  PN,  and  that 
PiM  is  greater  than  PiN. 

We  can  prove  PM  equal  to  PN  if  we  can  prove  triangles 
MKP  and  KNP  congruent.     This  the  student  can  readil 
do. 

We  can  prove  PiM  greater  than  PiN  by  drawing  the  line 
RN  and  noticing  that  NR  +  RP\,  which  is  equal  to  MR  + 
RPi,  is  greater  than  P,N.  (See  Art.  44.) 

Give  these  proofs  and  state  theorem  proved. 

Exercises. 

1.  Apply  this  theorem  to  show  that  the  perpendicular 
bisector  of  a  chord  of  a  circle  is  a  diameter. 

2.  Show  that  either  diagonal  of  a  rhombus,  produced,  is 
the  locus  of  points  equidistant  from  two  of  the  vertices. 

3.  What  is  the  locus  of  the  vertex  of  an  isosceles  triangle, 
if  the  base  is  fixed  while  the  length  of  the  equal  sides 
varies? 


150 


PLANE  GEOMETRY 


[V,  §  174 


174.  Theorem  III.     The  locus  of  a  point  equidistant  from 
two  lines  is  the  bisectors  of  the  angles  formed  by  those  lines. 


Given  the  lines  A  B  and  MNy  and  the  lines  YYi  and 
bisecting  the  angles  NRA,  MRB  and  BRN,  ARM  re- 
spectively. 

To  prove  that  YYi  and  XXi  form  the  locus  of  points  equidistant 
from  AB  and  MN. 

Analysis.  To  prove  this  let  P  be  any  point  on  either  YY1 
or  XXi,  and  PI  be  any  point  not  on  these  lines.  We  must 
prove  that  point  P  is  equidistant  from  AB  and  MN,  and 
that  PI  is  not  equidistant  from  A  B  and  MN.  Draw  the 
perpendiculars  PK  and  PiQ  to  A  B,  and  perpendiculars  PH 
and  PiT  to  MN.  The  lines  PK  and  PH,  which  represent 
the  distances  of  P  from  the  line  A  B  and  MN,  respectively, 
can  be  proved  equal,  if  we  can  prove  triangles  R  KP  and  RPH 
congruent.  The  student  can  readily  prove  this.  We  must 
now  prove  that  PiQ  does  not  equal  PI  T.  Since  PI  is  not  on 
the  bisectors,  PiQ  or  PI  T  will  be  cut  by  one  of  the  bisectors. 
Let  XXl  cut  Pi  T  at  point  J.  Draw  JF  _L  A  B.  Draw  PiF. 
Since  JT  is  equal  to  JF,  we  can  prove  P\T  greater  than  PiQ 
if  we  can  prove  that  PXJ  +  JF  is  greater  than  PiF,  which 
in  turn  is  greater  than  PiQ. 


Proof. 


PiF  >  PiQ. 
JF  >  P,F 
JT  =  JF. 
PiJ  +  JT  >  P 


Why? 
Why? 
Why? 
Why? 


v,  §  175]  LOCI  151 

PiJ  +  JT  >  P,Q.  Why? 

That  is  PiT  >  PjQ.  Why? 

State  the  theorem  that  we  have  just  proved. 

175.  Theorem  IV.     The  locus  of  a  point  equidistant  from 
two  given  parallel  lines  is  the  parallel  line  mid-way  between 
them. 

Let  the  student  give  figure,  analysis,  and  proof. 

Note.  In  the  figure  of  Theorem  III  imagine  the  points 
H  and  K  remaining  fixed  while  the  point  of  intersection  R 
moves  off  farther  and  farther  to  a  very  great  distance  to  the 
left.  Line  YYi  moves  with  R,  but  line  XXi  keeps  its  posi- 
tion, while  the  lines  BA  and  MN  approach  the  condition  of 
being  parallel  to  XX\. 

176.  Theorem  V.     The  locus  of  a  point  at  a  given  distance 
from  a  given  line  consists  of  two  lines  parallel  to  the  given  line, 
one  on  either  side  of  it,  and  at  the  given  distance  from  it. 

Prove  by  marking  a  point  on  one  of  the  parallel  lines  and 
showing  that  it  is  at  the  given  distance  from  the  given  line, 
then  marking  a  point  not  on  one  of  the  parallels  and  showing 
that  it  is  not  at  the  given  distance  from  the  given  line. 

Exercises. 

1.  Draw  two  intersecting  lines,  and  construct  the  locus  of 
points  equidistant  from  these  lines. 

2.  Draw  two  parallel  lines  three  inches  apart.     Construct 
the  locus  of  points  equidistant  from  these  lines. 

3.  Construct  the  locus  of  points  one  inch  from  a  given  line. 
Do  this  by  drawing  two  perpendiculars  to  the  given  line,  and 
marking  on  them  points  one  inch  from  the  given  line. 

177.  In  the  following  it  will  be  taken  for  granted  that  the 
work  is  all  in  the  same  plane.     Then  only  two  conditions  are 
necessary  to  fix  a  point  completely.     If  only  one  condition 
be  given,  the  point  is  left  free  to  move  and  describe  a  locus. 
To  make  this  clear  consider  the  following  problem. 


152  PLANE  GEOMETRY  [V,  §  178 

178.  Problem  in  Loci.     To  find  a  point  r  units  distant  from 
a  given  point,  and  d  units  distant  from  a  given  line. 


We  have  given  as  standards  of  reference: 

1st.        point  A, 

2nd.       line  MN. 

We  have  given  these  conditions  to  be  satisfied : 

1st.        The  point  we  are  locating  is  r  units  from  point  A . 

2nd.       The  point  we  are  locating  is  d  units  from  line  MN. 

Analysis.  The  locus  of  a  point  which  satisfies  the  first 
condition  is  a  circumference  about  A  as  a  center,  with  a  ra- 
dius r.  Why? 

The  locus  of  a  point  which  satisfies  the  second  condition, 
is  two  lines  one  on  each  side  of  line  MN  at  the  distance  d 
from  MN.  Call  these  lines  XXl  and  YYi. 

Where  circumference  C  cuts  the  lines  XXi  and  YY1}  PI, 
P2,  P3,  P4,  are  the  required  points.  Each  one  of  these  points 
satisfies  the  required  conditions. 

Let  the  student  give  construction  and  proof. 

Discussion.  There  is  nothing  in  the  problem  stating  the 
relative  position  of  the  point  A  and  the  line  MN.  They  can 
be  placed  as  near  together  or  as  far  apart  as  we  choose.  Also 
r  and  d  may  be  any  lengths.  Make  drawings  placing  line 
MN  at  different  distances  from  point  A,  leaving  r  and  d  the 
same,  and  discover  that  there  may  be  one,  two,  three,  four 
points,  or  no  point  that  satisfies  these  conditions. 


v,  §  179]  LOCI  153 

179.  Problems  and  Theorems.     According  to  the  illus- 
tration above  solve  the  following  problems. 

1.  A  railroad  and  a  county  road  cross.     A  station  is  on  the 
railroad.     A  man  wished  to  place  a  saw-mill  equally  distant 
from  the  railroad  and  the  county  road,  and  a  quarter  of  a  mile 
from  the  station.     Find  the  possibilities  of  such  a  location. 

2.  In  placing  a  statue  on  a  public  square,  it  is  desired  to 
have  it  equidistant  from  two  sides  opposite  each  other,  and 
ten  feet  from  a  third  side.     Find  possibility  of  its  location. 

3.  A  woman  wished  to  plant  a  shrub  equidistant  from  the 
corner  of  her  house  and  the  corner  of  her  lot,  and  at  the  same 
time  equidistant  from  the  front  line  of  her  house  and  the  front 
line  of  her  lot.     Find  the  possibilities  of  its  location. 

4.  Draw  a  triangle  ABC.     Construct  the  perpendicular 
bisector  of  side  AB.     Of  what  points  is  this  the  locus?    Con- 
struct the  perpendicular  bisector  of  side  BC.     Of  what  points 
is  this  the  locus?     Do  these  two  bisectors  intersect?     Why? 


How  does  this  point  of  intersection  lie  with  reference  to  the 
points  C  and  A!  Therefore,  in  what  line  must  it  lie?  Why? 
What  can  you  now  say  about  the  perpendicular  bisectors  of 
the  sides  of  a  triangle?  What  circle  can  you  describe  with 
the  point  of  intersection  as  a  center? 

The  result  of  this  exercise  may  be  stated: 

Theorem  VI.  The  perpendicular  bisectors  of  the  sides  of  a 
triangle  meet  in  a  point  which  is  equidistant  from  its  vertices. 
This  point  is  the  center  of  the  circumscribed  circle. 


154 


PLANE  GEOMETRY 


[V,  §  179 


5.  Draw  a  triangle  ABC.  Construct  the  locus  of  points 
equidistant  from  sides  A B  and  AC.  (Do  not  forget  that  it 
requires  two  lines  for  this  locus.)  Construct  the  locus  of  a 
point  equidistant  from  sides  B  A  and  BC.  Do  these  lines 
intersect?  Why?  How  many  points  of  intersection  are 
there?  How  do  these  points  lie  with  reference  to  the  lines 
CA  and  C5?  Why?  On  what  other  lines  do  these  points 
fall?  What  can  you  say  about  the  bisectors  of  the  angles  of 
a  triangle? 


Co, 


Theorem  VII.  The  bisectors  of  the  angles  of  a  triangle  meet 
in  a  point  equidistant  from  its  sides.  This  point  is  the  center 
of  the  inscribed  circle. 

If  the  exterior  angles  are  bisected  also,  three  other  points 
equidistant  from  the  sides  are  located.  These  are  the  centers  of 
the  escribed  circles. 

6.  To  find  the  center  of  a  given  circle.  (Use  Theorem  VI.) 
It  will  of  course  be  sufficient  to  draw  perpendicular  bisec- 
tors of  any  two  sides  of  the  inscribed  triangle,  or  of  any  two 
chords.  Draw  a  circle  with  a  radius  of  two  inches  and  erect 
perpendicular  bisectors  of  several  chords. 


v,  §180]  LOCI  155 

7.  To  find  the  center  of  a  given  arc.     Draw  three  inter- 
secting circles  whose  centers  are  on  the 

given  arc,  and  draw  their  common 
chords.  They  will  intersect  at  the  cen- 
ter of  the  arc.  Prove. 

8.  A  design  for  wall  paper  is  a  cir- 
cle  inscribed   in  an  equilateral  trian- 

gle.     Make  such  a  design,  filling  a  square  piece  of  paper. 

9.  Make  a  design  in  which  a  circle  circumscribes  an  equi- 
lateral triangle. 

10.  Make  a  design  in  which  a  circle  circumscribes  a  square. 

180.  Exercises.  Location  of  circles  that  must  fulfill  given 
conditions. 

When  called  upon  to  find  the  locus  of  a  point  that  fulfills 
certain  conditions,  it  is  well  to  locate  several  points  that  ful- 
fill the  condition  and  draw  a  line  through  them.  If  you  can 
show  that  every  point  on  the  line  thus  determined  fulfills 
the  condition  and  that  there  exists  no  other  point  that  does 
do  so,  then  the  line  is  the  correct  locus. 

The  student  is  warned  against  failure  to  get  all  the  lines 
when  there  is  more  than  one,  as  in  the  case  of  the  locus  of  a 
point  equidistant  from  two  intersecting  lines  or  the  locus  of  a 
point  at  a  given  distance  from  a  given  line. 

1.  What  is  the  locus  of  the  center  of  a  circle  whose  radius  is  r : 

(a)  that  will  touch  a  given  straight  line  MN*? 

(b)  that  will  touch  a  given  circle  C? 

(c)  that  will  pass  through  a  given  point  P? 

2.  Construct  a  circle  that  will  touch  a  given  line  MN, 
and  pass  through  a  given  point  P.     Draw  several  such  cir- 
cles and  study  the  locus  of  their  centers. 

3.  Construct  a  circle  with  given  radius  that  will  touch  a 
given  line  MN  and  a  given  circle  C. 

4.  What  is  the  locus  of  the  center  of  a  circle  that  will  touch 
two  given  intersecting  lines? 


156  PLANE  GEOMETRY  [V,  §  iso 

6.  Construct  the  locus  of  the  center  of  a  circle  that  will 
touch  two  given  parallel  lines. 

6.  What  is  the  locus  of  the  center  of  a  circle  that  will 
touch  two  given  concentric  circles? 

7.  What  is  the  locus  of  the  center 
of  a  circle  that  will  pass  through 
two  given  points? 

8.  Construct  a   circle   that  will 
pass  through  three  given  points  not 
all  in  the  same  straight  line. 

9.  Construct  a  circle  with  given 
radius,  tangent  to  two  given  circles. 

10.  Locate  a  point  which  is  at  a 

given  distance  from  two  intersecting  circles.     Is  there  more 
than  one  such  point? 

11.  Cut  a  circle  of  two  inches  radius  out  of  one  corner  of  a 
sheet  of  paper  without  wasting  any  more  paper  than  necessary. 

12.  With  a  given  radius,  say  half  an  inch,  construct  as 
many  circles  as  possible  on  an  ordinary  sheet  of  paper,  with- 
out overlapping. 

13.  Construct  three  equal  circles  within  an   equilateral 
triangle  that  will  be  tangent  to  one  another  and  to  the  sides 
of  the  triangle.     (Each  circle  is  tangent  to  two  sides.)    This 
is  a  common  design. 

14.  What  is  the  locus  of  the  center  of  a  circle  that  will  be 
tangent  to  a  given  line  at  a  given  point? 

15.  What  is  the  locus  of  the  center  of  a  circle  that  will  be 
tangent  to  a  given  circle  at  a  given  point? 

Suggestion.  Draw  a  tangent  at  the  given  point  and  find 
the  locus  of  the  center  of  the  circle  that  will  be  tangent  to  this 
tangent  at  the  given  point. 

16.  Construct  a  circle  that  will  be  tangent  to  the  line  AB  at 
the  point  A,  and  pass  through  a  point  C  not  on  the  line. 

17.  In  Exercise  16  how  will  any  angle  inscribed  in  the  seg- 
ments of  the  circle  constructed  compare  with  the  angle  CAB? 


V,  §  180] 


LOCI 


157 


18.  Construct  a  circle  that  will  be  tangent  to  a  given  semi- 
circle  and  its   diameter.     This   is   a   common 

design. 

19.  Construct  a  circle  which  shall  be  tangent 

to  one  arm  of  a  given  angle  at  the  vertex  and  cut  a  given 
segment  from  the  other  arm. 

20.  If  two  circles  intersect,  draw  two  circles  that  will  be 
tangent  to  both  of  them,  one  tangent  internally  and  the  other 
externally. 

21.  The  adjoining  figure  consists 
of  three  Gothic  arches  and  a  tangent 
circle. 

Construct  such  a  figure  taking 
AC  =  CB  =  2  inches. 

Suggestion.  To  draw  the  tangent 
circle  first  locate  the  point  0  as  follows : 

(a)  Point  0  must  lie  on  line  CD. 

(b)  Point  0  must  lie  on  arc  EO  which  has  the  same  center 
as  arcs  CF  and  BD,  and  is  mid- way  between  them. 

Show  that  r  =  JA5. 

22.  What  is  the  locus  of  the  vetex  of  a  triangle  having  a 
fixed  base  and  a  given  altitude? 

23.  What  is  the  locus  of  the  vertex  of  a  triangle  having  a 
fixed  base  and  a  given  median  to  the  base? 

24.  What  is  the  locus  of  the  vertex  of  a  triangle  having  a 
given  base  and  a  given  area? 

25.  What  is  the  locus  of  the  vertex  of  a  triangle  having  a 
given  base  and  a  given  angle  opposite  it? 

26.  Construct  a  triangle  having  a  given  base,  a  given 
altitude,  and  a  given  angle  at  the  base. 

27.  What  is  the  locus  of  the  mid-points  of  chords  drawn 
parallel  to  each  other  in  a  given  circle? 

28.  What  is  the  locus  of  the  mid-points  of  chords  drawn 
through  a  given  point  on  a  given  circumference? 


158 


PLANE  GEOMETRY 


[V,  §  181 


181.  Theorem  VIII.  The  locus  of  the  vertex  of  a  given 
angle  whose  arms  pass  through  two  given  fixed  points  is  the  arc 
capable  of  containing  the  angle  and  passing  through  the  given 
points.  See  Art.  155,  Corollary. 


182.  Problem  I.     To  construct  an  arc  on  a  given  line  as  a 
chord,  capable  of  containing  a  given  angle. 


Given  the  angle  ABC  and  the  segment  m. 

To  construct  an  arc  on  m  as  a  chord  capable  of  containing 
angle  ABC. 

Analysis.  Suppose  that  the  work  is  done  (see  figure)  and 
the  chord  AC  is  equal  to  the  given  line  m  and  angle  ABC  is 
equal  to  the  given  angle  ABC. 

Examining  the  figure  we  find  that  if  Z  ACR  is  formed  by 
drawing  tangent  CR,  Z  ACR  =  Z.  ABC.  Why?  Then  if 


v,  §  183]  LOCI  159 

we  construct  an  arc  of  which  AC  is  the  chord  and  CR  is  a 
tangent  making  Z  ACR  =  Z  ABC,  this  arc  will  contain 
Z  ABC,  and  is  the  arc  required. 

Therefore  our  problem  resolves  itself  into  the  problem  of 
constructing  an  arc  which  shall  have  AC  as  a  chord  and  CR 
as  a  tangent. 

This  in  turn  becomes  the  problem  of  finding  the  center  of  a 
circle,  which  shall  pass  through  the  given  points  A  and  C  and 
touch  the  given  line  CR  at  the  point  C,  Z  ACR  having  been 
made  equal  to  the  given  angle  ABC. 

The  locus  of  the  center  is 

1st.  the  perpendicular  bisector  of  the  line  joining  A 
andC. 

2nd.  the  line  perpendicular  to  the  line  CR  at  the 
point  C. 

The  point  of  intersection  of  these  two  loci  is  the  point  re- 
quired, that  is,  the  center  of  the  arc  whose  chord  is  line  m  and 
capable  of  containing  angle  A  BC. 

Make  this  construction  and  prove  correctness  of  the 
process. 

183.  Exercises. 

1.  Construct  a  triangle  having  given  the  base,  the  altitude, 
and  the  angle  opposite  the  base. 

2.  Construct  a  triangle  having  given  the  base,  the  angle 
opposite  the  base,  and  the  median  to  the  middle  of  the 
base. 

3.  Construct  a  triangle  having  given  the  base,  altitude, 
and  median  to  the  base. 

4.  Given  the  hypotenuse  of  a  right  triangle,  what  is  the 
locus  of  the  vertex  of  the  right  angle? 

5.  If  several  triangles  all  have  the  same  base,  and  equal 
angles  opposite  the  base,  prove  that  the  bisectors  of  these 
angles  all  pass  through  the  same  point. 


160 


PLANE  GEOMETRY 


[V,  §  184 


PART  II— COORDINATE  GEOMETRY 

184.  Coordinates.  We  shall  now  study  a  very  convenient 
method  for  fixing  positions  on  a  plane,  and  shall  then  make 
use  of  the  method  to  demonstrate  some  familiar  problems. 
We  shall  also  add  some  new  theorems  and  problems. 

To  locate  the  position  of  a  point  on  a  sheet  of  paper  or 
any  plane  surface,  we  follow  the  same  plan  as  is  used  to  locate 
a  place  on  the  earth's  surface.  Draw  two  lines  of  reference 
which  cut  at  right  angles;  one  extending  from  left  to  right 
and  the  other  up  and  down.  (Always  use  cross-section  paper 
for  this  work.) 

The  plane  of  the  paper  is  now  divided  into  four  quarters, 
called  quadrants.  They  are  numbered  I,  II,  III,  IV  as  shown 
in  the  figure.  We  now  locate  a  point  by  giving  its  distance 
from  the  two  reference  lines,  stating  also  whether  these  dis- 
tances are  measured  to  the  right  or  left,  upward  or  down- 
ward. The  line  x'x,  as  shown  in 
the  figure,  is  called  the  axis  of  the 
abscissa.  The  line  y'y  is  called  the 
axis  of  the  ordinate.  Distances 
upward  from  the  axis  of  the  ab- 
scissa are  positive ;  downward,  nega- 
tive. Distances  to  the  right  of  the 
axis  of  the  ordinate  are  positive; 
to  the  left,  negative.  The  point 
of  intersection  marked  0  is  called 
the  origin.  The  distance  which  a  point  is  to  the  right  or 
left  of  the  axis  of  the  ordinate,  measured  in  the  direc- 
tion of  the  axis  of  the  abscissa,  is  called  the  abscissa  of  the 
point.  The  distance  which  a  point  is  from  the  axis  of  the 
abscissa,  measured  in  the  direction  of  the  axis  of  the  ordinate, 
is  called  the  ordinate  of  the  point.  The  two  together  are 
called  the  coordinates  of  the  point  which  is  fixed  by  them. 


y 


V,  §  185] 


COORDINATE  GEOMETRY 


161 


(2,5) 


We  usually  represent  the  abscissa  of  a  point  by  x,  and  its 
ordinate  by  y;  then  the  two  numbers  x,  y  are  the  coordinates 
of  the  point.  However,  any  other  letters  may  be  used. 

Example.  Suppose  we  wish  to  locate  a  point  for  which  x  =  2  and 
y  =  5,  that  is,  the  abscissa  is  2  and  the  ordinate  is  5. 

Since  the  abscissa  is  2,  the  point  lies  2  units  to  the 
right  of  the  axis  of  the  ordinate.  It  is  therefore  on 
a  line  drawn  2  units  to  the  right  of  the  axis  of  the 
ordinate  and  parallel  to  that  axis.  This  line  we 
designate  by  the  equation  x  =  2,  because  this  equa- 
tion is  true  for  any  point  on  the  line,  and  is  true 
for  no  other  point.  This  line  is  the  locus  of  the 
point  whose  abscissa  is  2. 

Since  the  ordinate  of  the  point  is  5,  the  point 
must  be  somewhere  on  a  line  drawn  parallel  to  the 
axis  of  the  abscissa  and  5  units  above  it.  This  line  is  designated  by 
the  equation  y  =  5.  Why?  This  line  is  the  locus  of  a  point  whose 
ordinate  is  5. 

The  required  point  must  be  at  the  intersection  of  the  loci  just  drawn. 
We  call  it  the  point  (2,  5) ;  here  we  inclose  in  parentheses  the  coordinate 
of  the  point,  writing  first  the  abscissa,  and  then  the  ordinate,  with  a 
comma  between. 

A  point  whose  abscissa  is  x  and  whose  ordinate  is  y  is  designated  by 
the  symbol  (x,  y}. 

185.  Exercises.  Draw  the  lines  on  which  each  of  the  fol- 
lowing points  must  lie.  On  each  line  write  the  equation 
which  describes  it.  As  has  just  been  explained  such  a  line 
is  the  locus  of  the  point  which  lies  on  it. 

1.  What  is  the  locus  of  a  point  whose  abscissa  is  3?    Whose 
ordinate  is  7?     What  point  is  at  the  intersection  of  the  two 
loci?     Draw  a  complete  figure  as  shown  above,  marking  in 
the  equations  of  the  lines  and  the  symbol  for  their  point  of 
intersection. 

2.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  3  and  the 
ordinate  is  —  7. 

3.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  5  and 
the  ordinate  is  2. 


162 


PLANE  GEOMETRY 


[V,  §  186 


4.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  5  and 
the  ordinate  is  —  2. 

5.  Proceed  as  in  Exercise  1,  when  the  abscissa  is  —  3  and 
the  ordinate  is  —  3. 

6.  Make  diagrams  as  in  Exercise  1  for  each  of  the  follow- 
ing points: 

(2,  6);  (-  5,  4);  (6,  -  9);  (12,  -  7);  (-  2,  -  3);  (0,  -  9); 
(2,0). 

7.  What  line  is  described  by  the  equation  x  =  0?     What 
line  is  described  by  the  equation  y  =  0?     What  point  lies  at 
the  intersection  of  these  loci? 

8.  Locate  the  points  (9,  4);  (-  3,  -  1);  (4,  -  3).     Join 
by  straight  lines.     W^hat  kind  of  a  figure  is  formed? 


/ 


(2,3) 


186.  The  Distance  between  two 
Given  Points.  We  wish  to  express 
the  distance  between  two  given 
points  when  the   coordinates  of 
these  points  are  known.     For  this 
we  locate  the  points  as  in  the  ad- 
joining figure  and  by  counting 
and  using  the  Pythagorean  Theo- 
rem we  find  the  distance. 

187.  Exercises. 

1.  Determine  the  distance  between  each  of  the  following 
pairs  of  points: 

(1,2),  (4,6). 
(3,2),  (7,5). 
(2,1),  (4,3). 
(-2,1),  (1,3). 
(-33, -2),  (1,1). 
(-6,2),  (-6,7). 
(3,  -  1),  (-  1,  -  3). 
(5,2),  (-7, -3). 


(a) 
(b) 
(c) 
(d) 
(e) 
(f) 

(g) 
(h) 


V,  §  188] 


COORDINATE  GEOMETRY 


163 


2.  Determine  the  lengths  of  the  sides  of  the  triangles 
shown  in  each  of  the  following  figures. 


Y/V 


\ 


V 


3.  Show  that  the  points  (2,  2),  (4,  6),  and  (-  3,  7)  form  an 
isosceles  triangle.     Draw  an  accurate  figure. 

4.  Show  that  the  distance  from  the  origin  (0,  0)  to  any 
point  (a,  6)  is  \/a2  +  62. 

6.  Draw  the  quadrilateral  whose  vertices  are  (6,  6),  (—  2, 
4),  (-  6,  -  2)  and  (-  4,  -  4).  Calculate  the  lengths  of  its 
sides  and  its  diagonals. 

188.  The  Point  Mid-way  between  Two  Given  Points. 

Study  each  of  the  following  figures  to  determine  the  position 
of  point  Pj  which  is  mid-way  between  the  given  points  P\ 
and  PZ. 


164  PLANE  GEOMETRY  [V,  §  189 

Notice  that  the  right  triangle  PiHP  is  equal  to  the  right 
triangle  PKP2.     Why? 

Therefore    P^H  =  J  P^L  =  J  (7  - 1)  =  3.     Why?     Art.  84. 
Similarly      HP  =  J  LP2  =  \  (5  - 1)  =  2. 
Hence         abscissa  of  P  =  abscissa  of  PI + PiH  =  1+3=4, 

and         ordinate  of  P  =ordinate  of  Pi+HP  =  1+2  =3. 

Therefore  in  the  first  figure  P  is  the  point  (4,  3),  and  in  the 
second  figure  P  is  the  point  (1,  1). 

189.  Exercises. 

1.  In  the  first  figure  of  Exercise  2,  Art.  187,  find  the  mid- 
point of  two  sides  of  the  triangle.     Join  these  points  by  a  line 
and  find  the  length  of  the  line.     Compare  the  result  with  the 
length  of  the  third  side.     State  a  theorem  of  geometry  which 
covers  this  case. 

2.  Locate  the  points  (2,  -  7),  (5,  -  3),  (-  9,  -  2).    Draw 
the  triangle  and  find  the  lengths  of  its  three  medians. 

3.  Locate  the  points  (9,  6),  (-  2,  3),  (-  4,  -  4),  (7,  -  1). 
Show  that  the  opposite  sides  are  equal  and  hence  that  the 
figure  is  a  parallelogram.     Find  the  mid-point  of  each  diag- 
onal and  compare  results.     What  theorem  covers  this  case? 

4.  Locate  the  points  (-  4,  -  1),  (7,  -  5),  (3,  3).     Add  a 
fourth  point  so  as  to  form  a  parallelogram.     What  are  its 
coordinates?     Prove  that  your  figure  is  a  parallelogram. 

5.  In  the  quadrilateral  of  Exercise  5,  Art.  187,  find  the 
mid-point  of  each  side.     Join  these  one  after  the  other  by 
straight  lines.     Show  that  the  new  figure  is  a  parallelogram 
by  two  methods. 

First :  Show  that  the  opposite  sides  are  equal. 
Second :  Show  that  the  diagonals  bisect  each  other  by  show- 
ing that  the  mid- points  coincide. 

6.  In  the  parallelogram  of  Exercise  4  find  the  mid-point 
of  each  side.     Join  the  opposite  mid-points  by  straight  lines 
and  find  the  lengths  of  these  lines.     Show  that  they  bisect 
each  other  by  showing  that  their  mid-points  coincide. 


V,  §  190] 


COORDINATE  GEOMETRY 


165 


190.  The  Slope  of  the  Line  through  two  Given  Points. 

The  slope  of  a  line  is  a  number 
used  to  indicate  the  direction  of 
the  line.  In  the  adjoining  figure 
let  the  line  pass  through  points  PI 
and  P2. 

Then  the  slope  of  the  line  is  de- 
fined to  be  the  value  of  the  frac- 


tion 


Slope  of  line  PiP2  = 


MP2 
PiM' 


y 


The  slope  of  a  line  may  be  either  positive  or  negative,  as 
illustrated  below. 


\P2 


Slope 


2-(-l) 


4 2 

Slope  =  — ^ — ^       Slope  = 


6-1  -3-3 

=  3  _  1  _7 

5'  3*  3' 

In  a  problem   either  point  may  be  called  PI  and  the 
other  P2. 

When  the  line  leans  to  the  right  the  slope  is  positive.    When 
it  leans  to  the  left  the  slope  is  negative. 
Lines  which  have  equal  slopes  are  parallel. 

191.  Exercise. 

1.  Write  the  slope  of  the  line  through  each  pair  of  points 
in  Exercise  1,  Art.  187. 


166  PLANE  GEOMETRY  [V,  §  192 

2.  Locate  points  (9,  6),  (-  2,  3),  (-  4,  --  4),  (7,  --  1). 
Join  in  order  by  straight  lines.     Show  that  the  opposite  sides 
of  the  figure  have  equal  slopes,  and  hence  are  parallel.    What 
kind  of  a  figure  is  formed? 

3.  Locate  the  points  (5,  4),  (—  3,  2),  (1,—  6)  and  join  them 
by  straight  lines,  forming  a  triangle. 

Determine  the  coordinates  of  the  mid-points  of  any  two 
sides  of  the  triangle.  Then  show  that  the  line  joining  these 
mid-points  is  parallel  to  the  third  side. 

4.  Construct  a  quadrilateral  whose  vertices  are  the  points 
(4,  7),   (-  2,  5),   (-  6,  -  3),   (6,  -  5).      Find  the  coordi- 
nates of  the  mid-points  of  the  four  sides.     Show  that  the 
mid-points  are  the  vertices  of  a  parallelogram,  by  showing 
that  the  opposite  sides  of  the  figure  formed  by  them  are 
parallel. 

192.  Summary  of  Chapter  V.    Loci  of  Points. 

Part  I — Loci  in  General. 

General  Discussion. 

Theorem  I.  The  locus  of  a  point  at  a  given  distance  from  a  given 
fixed  point  is  the  circumference  described  about  the  fixed  point  as  a 
center,  with  a  radius  equal  to  the  given  radius. 

Theorem  II.  The  locus  of  a  point  equidistant  from  two  given  fixed 
points  is  the  perpendicular  bisector  of  the  line  joining  them. 

Theorem  III.  The  locus  of  a  point  equidistant  from  two  given  inter- 
secting lines  consists  of  the  bisectors  of  their  included  angles. 

Theorem  IV.  The  locus  of  a  point  equidistant  from  two  given  parallel 
lines  is  a  line  parallel  to  them  and  mid-way  between  them. 

Theorem  V.  The  locus  of  a  point  at  a  given  distance  from  a  given 
line  consists  of  a  pair  of  parallels  one  on  each  side  of  the  given  line  at 
the  given  distance  from  it. 

Locations  of  points  by  means  of  loci. 

Discussion  and  Examples. 

Theorem  VI.  The  perpendicular  bisectors  of  the  sides  of  a  triargle 
meet  in  a  point  which  is  equidistant  from  the  vertices  of  the  triangle. 

Theorem  VII.  The  bisectors  of  the  angles  of  a  triangle  meet  in  a  point 
which  is  equidistant  from  the  sides  of  the  triangle. 


v,  §  192]  COORDINATE  GEOMETRY  167 

Theorem  VIII.  The  locus  of  the  vertex  of  a  given  angle  whose  arms 
pass  through  two  fixed  points,  is  the  arc  capable  of  containing  the 
angle,  and  passing  through  the  given  points. 

Problem  I.  On  a  given  line  segment  as  a  chord  to  construct  the  seg- 
ment of  a  circle  capable  of  containing  a  given  angle. 

Part  II — Coordinate  Geometry. 

Definitions.     Location  of  points  on  a  plane. 
Finding  distances  between  two  points. 
Finding  a  point  mid-way  between  two  points. 
Finding  the  slope  of  a  line  between  two  points. 

1.  Into  what  two  parts  is  this  chapter  divided? 

2.  What  is  meant  by  the  locus  of  a  point  to  fulfil  a  given  condition? 

3.  If  you  wish  to  locate  a  point  in  a  plane  how  many  conditions  must 
be  given?     Hence  how  many  loci  will  be  necessary  to  fix  the  point? 

4.  When  you  have  given  the  abscissa  and  ordinate  of  a  point,  have 
you  two  conditions  for  its  location  given?    What  are  the  loci  that  cor- 
respond to  them? 

6.  Given  the  abscissas  and  the  ordinates  of  two  points,  how  do  you 
find  the  length  of  the  line  joining  them?  How  do  you  find  the  mid- 
point of  this  line? 

6.  Have  any  of  the  theorems  of  this  chapter  been  proved  without 
any  of  the  preceding  chapters?     If  so,  state  them. 

7.  With  what  preceding  theorems  are  the  theorems  of  this  chapter 
most  closely  connected? 

Historical  Note.  The  subject  of  Coordinate  Geometry  is  often  termed 
"Cartesian"  Geometry,  after  the  name  of  its  founder,  Descartes  (1596- 
1650).  He  introduced  the  idea  of  representing  the  position  of  a  point 
by  means  of  coordinates,  and  the  locus  of  a  point  by  means  of  an  equa- 
tion between  those  coordinates.  The  resulting  union  of  algebra  and 
geometry  has  been  a  powerful  aid  in  the  development  of  both  subjects, 
and  constitutes  one  of  the  great  steps  in  the  development  of  mathe- 
matical science. 


CHAPTER  VI 

PART  I— RATIO  AND  PROPORTION.    PART  II— SIMI- 
LAR FIGURES.     PART  III— TRIGONOMETRIC 
FUNCTIONS 

PART  I— RATIO  AND  PROPORTION 

193.  Definitions  and  Notation.  We  have  been  dealing  with 
the  subject  of  ratio  from  the  beginning  of  our  course.  We 
have  defined  the  ratio  between  two  numbers  as  the  number 
by  which  the  second  must  be  multiplied  to  obtain  the  first. 

If  a  -:-  6  =  r,  then  r  is  the  ratio  of  a  to  b,  meaning  that 
br  =  a.  This  ratio  is  often  indicated  by  the  symbol  a  :  b}  or 

by  r,  in  place  of  a  -r-  b. 

If  four  numbers  are  so  related,  that  we  must  multiply  the 
fourth  by  the  same  number  to  obtain  the  third  that  we  must 
multiply  the  second  by  to  obtain  the  first,  the  four  numbers 
are  said  to  form  a  proportion.  In  other  words  a  proportion  is 
the  equality  of  two  ratios. 

Thus:  -r  =  -7  ,  or  a-f-  b  =  c  -r-  d,  or  a  :b  =  c  :  d,  are  sym- 
bols for  the  equality  of  ratios,  or  proportions. 

The  equality  of  more  than  two  ratios  is  called  a  continued 
proportion.  Thus : 

ace  7  j  /. 

5-  =  d  =  J=  ;  ma:b  =  c:d  =e:f  = 

Since  numbers  may  be  represented  by  line-segments,  the 
subject  of  ratio  and  proportion  is  treated  both  geometrically 
and  algebraically. 

168 


VI,  §  194]  RATIO  AND  PROPORTION  169 

We  note  the  following  correspondence  of  terms: 

Geometry  Algebra 

Line-segment  Number 

Commensurable  Rational 

Incommensurable  Irrational 

Line-segment  is  the  portion  of  a  line  between  two  definitely 
located  points  on  a  line. 

Number  expresses  the  ratio  which  such  a  segment  bears  to 
another  line-segment  selected  as  a  unit  of  measure. 

Determining  the  ratio  is  equivalent  to  measuring  the  line. 
The  expression  of  this  ratio  is  number. 

We  have  seen  that  some  line-segments  exist  for  which  a 
common  unit  of  measure  can  be  found.  These  line-segments 
are  said  to  be  commensurable.  Numbers  which  express  their 
ratios  are  called  rational  numbers. 

Other  line-segments,  such  as  the  diagonal  and  the  side  of  a 
square,  the  circumference  and  the  diameter  of  a  circle,  have 
no  common  unit  of  measure.  We  call  them  incommensurable 
lines.  Numbers  which  express  their  ratios  are  called  ir- 
rational. 

The  radical  sign  and  the  fractional  exponent  have  been 
introduced  in  the  expression  of  these  ratios.  (Review  First 
Course,  page  172.) 

In  practical  work  all  numbers  are  treated  as  commensur- 
able, since  we  can  make  the  unit  of  measure  as  small  as  we 
please,  and  so  have  the  work  as  nearly  accurate  as  necessary. 

194.  Definitions. 

The  terms  of  a  ratio  are  named  antecedent  and  consequent 
respectively. 

Thus,  if  r  is  the  ratio  of  a  to  b,  r  =  a  :  6,  then  a  is  the  ante- 
cedent and  b  is  the  consequent.  These  correspond  to  the 
numerator  and  the  denominator  of  a  fraction,  the  dividend 
and  the  divisor  of  a  division. 


170 


PLANE  GEOMETRY 


[VI,  §  195 


In  a  proportion  the  first  and  last  terms  are  called  extreme 
terms,  and  the  second  and  third  terms  are  called  mean  terms. 

Thus  in  the  proportion  2:3  =  4  :  6,  2  and  6  are  extremes 
and  3  and  4  are  means. 

When  a  mean  term  is  repeated,  it  is  called  a  mean  pro- 
portional. 

Thus  in  the  proportion  2  :  10  =  10  :  50,  10  is  the  mean 
proportional  between  2  and  50. 

Exercise.  Give  several  pairs  of  numbers  which  can  be 
used  as  means  when  the  extremes  are  4  and  16.  What  is 
the  mean  proportional  between  these  numbers? 

195.  Geometric  Illustrations  of  Ratio  and  Proportion. 

Since  a  rigid  treatment  of  the  theory  of  ratio  and  propor- 
tion is  too  difficult  for  beginners,  we  shall  regard  the  follow- 
ing geometric  treatment  more  as  illustrative,  and  thus  show 
the  meaning  of  the  algebra. 

Definitions. 

If  through  a  point  any  number  of  lines  are  drawn,  the  figure 
so  formed  is  called  a  pencil  of  lines.  The  point  is  called  the 
vertex  of  the  pencil. 

A  number  of  lines  parallel  to 
one  another  is  called  a  pencil  of 
parallels. 

Exercise.  Draw  a  pencil  of 
three  straight  lines  cut  by  two  parallel  lines  as  in  the  figure. 

Measure  a,  b,  c,  d,  e,  f.  v 

ace 

Calculate  the  ratios  T  ,  -j,  7-      Di- 
b   a  f 

vide  out  to  two  decimal  places. 

Do  you  find  any  relation  between 
these  ratios? 

Calculate  the  ratios  -,  -,  -. 
ace 


VI,  §  196]  RATIO  AND  PROPORTION  171 

Do  you  find  a  relation  between  these  ratios? 

If  you  divide  1  by  each  of  the  ratios  of  the  first  set,  you  will 
obtain  the  ratios  of  the  second  set.  The  ratios  of  one  set  are 
said  to  be  the  reciprocals  of  the  corresponding  ratios  of  the 
other. 

The  truth  brought  out  by  your  measurements  is  expressed 
by  the  following  theorem. 

196.  Theorem  I.  //  a  pencil  of  lines  is  cut  by  a  pencil  of 
parallels  the  corresponding  segments  are  in  proportion. 


Given  the  pencil  of  line  whose  vertex  is  V,  cut  by  the  paral- 
lels P  and  PI,  cutting  the  segments  a,  6,  c,  d,  e,  /,  respectively. 

To  prove  that  ^  =  ^=- 

Proof.  Assuming  that  the  segments  a  and  6  are  com- 
mensurable, let  u  be  a  unit  of  measure  common  to  them. 
Let  m  be  the  number  of  times  that  u  is  used  to  measure  a, 
and  m'  the  number  of  times  that  u  is  used  to  measure  b. 
Through  the  points  of  division  on  a  and  b,  draw  lines  parallel 
to  P.  These  lines  will  be  parallel  to  PI  also.  Why? 

These  lines  will  divide  c  and  e  into  m  equal  parts,  and  d 
and  /  into  m'  equal  parts?  Why?  Each  part  of  c  and 
d  is,  say,  HI  units  long,  and  each  part  of  e  and  /  is  u^ 
units  long. 


172  PLANE  GEOMETRY  [VI,  §  197 

So  we  have                 a  =  mu,  b  =  m'u', 

c  =  mui,  d  =  mfu\\ 

e  =  muz}  /  =  m'uz. 
Forming  our  ratios,  we  have  : 

a  _  mu  _  m  t 

b      m'u  m" 

c      mui  m 


_  _ 

d      m'u\      m" 

i  *        ™*   .  *  .  Why? 

/      mid     m 

Therefore  l  =  ^  =  ~f  Why? 

State  the  proposition  proved. 

Remark.  Notice  that  we  assumed  a  and  6  to  be  com- 
mensurable lines.  But  a  might  be  the  side  of  a  square  of 
which  b  is  the  diagonal,  in  which  case  the  proof  given  above 
would  not  answer.  However,  as  we  said  before,  we  shall 
assume  our  propositions  hold  for  incommensurable  cases. 
These  will  be  discussed  later  when  the  theory  of  limits  has 
been  studied.  Art.  289. 

The  proof  has  been  given  for  three  lines  of  the  pencil, 
leading  to  three  equal  ratios.  Evidently  any  number  of 
lines  could  be  used. 

197.  Definition.     The  reciprocal  of  a  number  is  1  ^  by 

1        2 
that  number  :  thus,  the  reciprocal  of  2  is  1  -f-  2  or  ^,  of  ^  is 

2i          o 

2  3     ,  a.   .       a       b 
1  -f-  -  or  x,  of  T  is  1  -!-  T  or  -. 

3  2'       b  b       a 

By  inverting  the  ratios  used  in  Theorem  I  we  could  have 
shown  just  as  well  that 

b  =  d=f 
ace' 

The  relation  of  this  result  to  the  result  of  Theorem  I  is 
usually  stated  thus: 


VI,  §  198]  RATIO  AND  PROPORTION  173 

198.  Theorem  II.     //  quantities  are  in  proportion,   they 
are  in  proportion  by  inversion. 

199.  Theorem  III.     //  four  quantities  are  in  proportion, 
they  are  in  proportion  by  alternation,  that  is,  the  first  is  to  the 
third  as  the  second  is  to  the  fourth. 

Geometric  proof: 

Taking  the  figure  and  notation  of  Theorem  I,  by  the  same 

plan  of  proof,  show  that  -  =  %  . 


Algebraic  proof: 
We  have 


I  =  j  • 
6      d 


Multiply  both  sides  of  the  equation  by  -  ; 
Therefore  ~c=\'  Why? 

200.  Exercise.  In  your  figure  for  the  preceding  exercise 
(Art.  195)  measure  VA,  VC,  VE,  BA,  DC,  FE.  Find  the 

VA    VC   VE 
*  BA'  DC'  FE' 

How  do  these  ratios  compare  with  one  another?  How  do 
they  compare  with  those  of  the  preceding  exercise?  Explain 
from  the  figure  why  each  ratio  of  this  set  exceeds  the  cor- 
responding ratio  in  the  first  set  by  1. 

Using  the  notation  of  Theorem  I,  VA  =  a  +  b,  VC  = 
c  +  d,  VE  =  e  +  /;  therefore  these  equal  ratios  may  be 
written 

a  +  b      c  +  d      e 


_ 
b  d  f 

Apply  this  to  each  of  the  following  proportions: 

(a)  3:5  =  6:10=12:20; 

(b)  6:4-30:20=12:8  =  3:2. 
Expressed  as  a  theorem,  we  have 


174  PLANE  GEOMETRY  [VI,  §  201 

201.  Theorem  IV.  In  a  series  of  equal  ratios,  the  sum  of 
the  first  antecedent  and  first  consequent  is  to  the  first  consequent 
as  the  sum  of  the  second  antecedent  and  second  consequent  is  to 
the  second  consequent,  as  the  sum  of  the  third  antecedent  and 
third  consequent  is  to  the  third  consequent,  and  so  on. 

Geometric  Proof. 

Using  the  figure  and  notation  of  Theorem  I,  we  have 

a  -f-  b  =  mu  +  m'u,    =  (m  +  mf)  u', 
c  +  d  =  mui  +  m'ui,  =  (m  +  m'}  u\; 
e  +/  =  muz  +  m'uz,  =  (m  +  m')  Uz. 

Then  our  ratios  are 

a  +  b     (m-\-m'}u     c  +  d  _(m  +  m')ui     e+f     (m  +  m')uz 


_ 

b  m'u  d  m'ui  f 

Therefore  we  have 

o  +  i  <,      e  +  d      e  + 


w 


b  d  f 

Algebraic  Proof. 
^.  ace 

Glven  6=d=7' 

We  may  add  1  to  each  ratio  without  destroying  the 
equality; 

so  we  have  |+l=|+l=^  +  l  Why? 

a+b      c+d      e+f 
Adding:  -^      -r      —. 

Explain  how  your  figure  shows  that  adding  1  to  the  first 
set  of  ratios  gives  the  second  set  of  ratios. 

Explain  the  changes  that  would  need  to  be  made  in  the 
proof  in  order  to  have  this  theorem  read: 

In  a  series  of  equal  ratios,  the  sum  of  the  first  antecedent 
and  consequent  is  to  the  first  antecedent,  as  the  sum  of  the 
second  antecedent  and  consequent  is  to  the  second  antecedent, 
and  so  071. 


VI,  §  202]  RATIO  AND  PROPORTION  175 

202.  Theorem  V.     In  a  series  of  equal  ratios,  the  difference 
between  the  first  antecedent  and  consequent  is  to  the  first  con- 
sequent as  the  difference  between  the  second  antecedent  and  con- 
sequent is  to  the  second  consequent,  as  the  difference  between  the 
third  antecedent  and  consequent  is  to  the  third  consequent. 

Show  this  true  by  measurement,  then  show  it  to  be  true 
by  geometric  proof.  Algebraically  it  may  be  stated: 

if       °  -  c  -  e-  then       a~b  -  °~d  -  e~f 

b  ~  d  ~f  b  d  f    ' 

Make  the  algebraic  proof  of  this  by  subtracting  1  from  each 
of  the  given  ratios. 

Re-word  the  proposition  so  that  the  word  antecedent  shall 
take  the  place  of  consequent  as  the  second  term  of  each  ratio, 
and  prove. 

203.  Theorem  VI.     In  a  series  of  equal  ratios,  the  sum  of  the 
first  antecedent  and  consequent  is  to  their  difference,  as  the  sum 
of  the  second  antecedent  and  consequent  is  to  their  difference,  and 
soon. 

Show  by  measurements  of  lines.  Prove  by  geometry. 
Write  in  algebraic  language  and  prove  by  algebra. 

Suggestion.  The  algebraic  proof  can  readily  be  made  by 
dividing  the  algebraic  expression  of  Theorem  IV  by  the 
algebraic  expression  of  Theorem  V. 

204.  Definitions.     The  operation  of  passing  from  T  =  -3 

a  +  b      c  -\-  d        .       n    i  /< 
to  — T—  = — -j —        is  called  "  composition  "; 

a  —  b      c  —  d 

to  — T —  =  — j—       is  called  "  division  " ; 

b  a 

t  a  +  b  _  c  -f  d       is  called  "  composition  and  divi- 

a  —  b  ~  c  —  d  sion." 

The  word  division  is  here  used  in  a  sense  entirely  different 
from  its  usual  meaning. 


176  PLANE  GEOMETRY  [VI,  §  205 

205.  Theorem  VII.  In  a  series  of  equal  ratios  the  sum  of  the 
antecedents  is  to  the  sum  of  the  consequents  as  any  antecedent  is 
to  its  consequent. 

n         f*        p 

Given  the  equal  ratios  :-=-=-  =  ••••. 

a  +  c  -f  e.  .          a       c       e 
To  prove  that  —    _,  \   ,      -  =  -  =  -=..... 


Proof.     Since  the  ratios  are  equal,  let  the  common  value 
be  r. 

r™  ace 

Then  j-r,      5  -  r,      j  =  r,  •    --. 

Or  a  =  br,     c  =  dr,     e  =  fr,  -  •  -  -  . 

Adding:  a  +  c  +  e  +  .  .  .  .  =  (b  +  d  +  /+....)  r. 

,„,       £  a  +  c  +  e  +  .  .  .  .  a      c      e 

Therefore  . 


Exercise.     Consider  any  number  of  equal  fractions: 

2  6      12  =20 

3  ==  9  ~  18  ~  30' 

Form  a  new  fraction  by  adding  together  several  or  all  the 
numerators  and  the  corresponding  denominators.  Show  that 
the  new  fraction  is  equal  to  f  . 

5  4-  x 
Apply  this  principle  to  show  that  if  the  fractions 

3-x 
and  5  —  —  are  equal,  then  each  of  them  must  equal  2. 

o  —  dx 

We  can  make  use  of  Theorem  I  to  divide  a  line-segment  in 
a  given  ratio. 

206.  Definition.  Given  a  line  segment  A  B,  which  is 
divided  into  two  parts  AC  and  CB  by  a  point  C.  The  ratio 
of  these  segments  is  AC  :  CB,  and  line  AB  is  said  to  be 
divided  internally  in  the  ratio  AC  :  CB.  Here  point  C  falls 
between  A  and  B. 


VI,  §  207]  RATIO  AND  PROPORTION  177 

Examples. 


8  A  C          6         A  C          B 

5         AC       6      2 

2         C£  ~  3  ~  1 

207.  Problem  I.     To  dm'de  a  line-segment  internally  in  a 
given  ratio. 


Given  the  line-segment  MN  and  the  segments  r  and  s. 
Jo  divide  the  line-segment  MN  internally  in  the  ratio  of  r  :  s. 

Analysis.  We  can  form  a  pencil  of  two  rays,  one  being  the 
line-segment  MN  and  the  other  an  indefinite  straight  line 
MY,  M  being  the  vertex  of  the  pencil.  On  MY  lay  off  MH 
equal  to  r,  and  add  to  it  HK  equal  to  s.  Join  N  and  K. 
Now  by  drawing  a  line  through  H  parallel  to  N  K,  have  the 
line-segment  divided  in  the  ratio  of  r  :  s. 

Construct  and  prove  that  MP  :  PN  =  r  :  s. 

Discussion.  Consider  the  case  where  the  vertex  of  the 
pencil  is  taken  at  N. 

208.  Definition.  Given  a  line-segment  A  B  and  a  point 
C  on  AB  produced.  Then  C  is  said  to  divide  AB  externally 
in  the  ratio  AC  :  CB. 


Notice  that  the  order  of  the  letters  in  the  ratio  is  exactly 
the  same  as  for  internal  division. 


178  PLANE  GEOMETRY  [VI,  §  209 

Examples. 


B        c  A 


=  =      =  3 

C£       2  C£  ~  3  ~  CB  ~  7' 

209.  Problem  II.  To  dm'de  a  given  line-segment  externally 
in  a  given  ratio. 


Y 

Given  line-segment  M  N  and  segments  r  and  s. 
To  divide  MN  externally  in  the  ratio  r  :  s. 

We  follow  the  same  plan  as  in  Problem  I,  the  only  difference 
being  that  H  K  =  s  is  laid  off  from  H  toward  M  instead  of 
from  H  toward  F.  We  now  join  K  to  N  as  before,  and 
then  through  H  draw  a  line  parallel  to  KN.  This  will  meet 
MN  produced  at  P,  which  is  the  required  point  of  external 
division. 

~  ,  M  P      MH      r 

Construct  and  prove  that 


S 

Discussion.  Consider  the  case  where  the  vertex  of  the 
pencil  is  taken  at  N. 

Those  who  have  studied  "  First  Course"  will  notice  the 
correspondence  between  laying  off  segments  r  and  s  on  MY 
and  the  addition  of  line  segments  as  explained  on  pages  54-57. 
External  division  corresponds  to  dividing  a  line  into  two 
segments,  one  positive  and  the  other  negative.  The  given 
segment  is  divided  into  two  parts  one  positive  and  the 
other  negative. 

To  illustrate  this,  let  the  student  divide  a  segment  of  8 
units  in  the  ratio  5  :  -  2;  one  of  12  units  in  the  ratio  4  :  -  7; 
one  of  \/21  units  in  the  ratio  10  :  -  3. 


VI,  §  210]  RATIO  AND  PROPORTION  179 

210.  Exercises. 

1.  By  measurement  determine  the  approximate  ratio  of 
division  of  the  segments  MN  in  Arts.  207  and  209. 

2.  Draw  a  segment  10  units  long;  divide  it  internally  in 
the  ratio  3:5;  externally  in  the  ratio  3  :  5. 

3.  Draw  a  line-segment  10  units  long,  and  a  square  4  units 
on  each  side.     Divide  the  segment  internally  and  externally 
in  the  ratio  of  side  :  diagonal  of  the  square. 

If  a  :  b  —  c  :  d,  prove  the  following  relations : 
a2  _  c?  a^_  c2 

b2  "  d2'  6'  ab  ~  dT 


b2               d2  '  a2  -  b2      c2  -  d2' 

8.  2  a  :  3  b  =  2  c  :  3  d.  9.  2a  +  36  :  6  =  2  c+  3d 

ma  _  nc  ma  +  n6       me  +  nd 

10.  7    —   j.  11.     -  T  =   —      ; . 

mo       nd  nb                 nd 

12.    -r^ =-j   =  -r^ =— :.  13. 


4  a  -  5  6      4  c  -  5  d'  b2  d2 


Va*  +  62       Vc2  +  d2  \/3a2+562 

14.     =    -      : .  15.     ; =    ; . 

0  d  b  d 

16.  a3  :  63  =  c3  :  d3.  17.  aw  :  6n  =  cn  :  dn. 

18.  a3  +  63  :  63  =  c3  +  d3  :  c/3. 

19.  a3  -  63  :  c3  -  d3  =  63  :  d3. 

20.  a2  +  a6  +  b2  :  c2  +  cd  +  d2  =  62  :  d2. 

21.  a2  -  ab  +  62  :  c2  -  cd  +  d2  =  b2  :  d2. 

22.  a2  +  ab  +  b2  :  a2  -  a&  +  62  =  c2  +  cd  +  d2  :c2  -  cd  +  d2. 

23.  Va2  +  a&  +  b2  :  Vc2  +  cd  +  d2  =  ^a3  -  63  :  v'c3  -  d3. 

(See  Exercises  19,  20.) 

rf  a       c       e      ,  3a  +  2c  +  5e      a 

24'  K  6_=  5  =  /'  sh°W  that_36  +  2^  +  5/  =  6'    - 

25.  If  V2  -  x  :  x  =  x  -  3  :  \/2  +  x,  find  the  value  of  x  and 
check. 

26.  If  Vx  -  2  -Vx  +  2  :  \/x  +  3  =  Vx  -  3  :  Vx  -  2  + 
vx  +  2,  find  the  value  of  x  and  check. 


180  PLANE  GEOMETRY  [VI,  §  211 

Going  back  to  our  figure  of  the  pencil  of  lines  cut  by 
parallels,  there  are  still  equal  ratios  which  we  have  not 
investigated. 

211.  Theorem  VIII.  //  a  pencil  of  lines  is  cut  by  a  pencil 
of  parallels,  the  segments  of  the  lines  of  the  pencil  (measured 
from  the  vertex}  are  proportional  to  the  corresponding  segments 
of  the  parallels. 


Given  the  pencil  of  lines  with  vertex  V.  VA  and  VB  are 
the  segments  cut  on  one  of  the  rays  by  the  parallels  P  and  PI. 
AC  and  BD  are  the  corresponding  segments  cut  on  the  par- 
allels. 

To  prove  that  ^  =  ^. 

Analysis.  Since  the  only  theorem  that  we  have  had  on 
this  subject  deals  with  ratios  between  segments  of  the  rays 
of  the  pencil  of  lines,  we  shall  form  a  pencil  such  that  VA, 
VB,  AC,  BD  shall  be  the  segments  of  its  rays,  or  equal  to 
segments  of  its  rays.  If  we  draw  a  line  through  A  parallel 
to  VD,  we  shall  have  such  a  pencil.  Point  B  is  the  vertex 
of  this  pencil.  A C  is  not  a  segment  of  a  ray  of  the  pencil, 
but  if  we  can  show  AC  equal  to  RD,  R  being  the  point  where 
the  line  drawn  through  A  cuts  BD,  we  shall  have  proved  our 
theorem. 

Proof.    Give  this  proof  in  full. 


VI,  §  212]  RATIO  AND  PROPORTION  181 

212.  Theorem  IX.     //  a  pencil  of  lines  is  cut  by  a  pencil  of 
parallels,  the  corresponding  segments  cut  on  the  parallels  are  in 
proportion. 

Draw  figure,  analyze,  and  prove  by  application  of  Theorem 
VIII. 

By  limiting  Theorem  I  to  two  rays  and  two  parallels  we 
have: 

213.  Theorem  X.     //  a  line  is  drawn  parallel  to  one  side 
of  a  triangle,  it  divides  the  other  two  sides  proportionally. 

The  converse  of  this  theorem  is: 

214.  Theorem  XI.     //  a  line  divides  two  sides  of  a  triangle 
proportionally,  it  is  parallel  to  the  third  side. 


Given  the  triangle  ABC  with  the  line  MN  cutting  the  side 

AP      AO 

AB  at  P  and  side  CA  at  Q,  forming  the  proportion  *=>-=  =  —$, 

L  Jj       (^/C 

To  prove  that  line  MN  is  parallel  to  side  BC. 

Analysis.  We  shall  follow  the  usual  plan  in  the  proof  of  a 
converse  theorem,  that  is,  we  shall  draw  through  point  P  a 
line  PR  parallel  to  side  BC,  and  prove  that  PR  coincides 
with  MN.  This  will  prove  our  theorem. 

Proof.  Prove  this  theorem  by  answering  the  following 
questions.  Since  PR  was  drawn  parallel  to  BC,  what  pro- 
portion exists?  Why?  What  proportion  exists  from  the 
assumption  of  the  theorem?  Why?  What  two  ratios  are 
equal  to  one  another,  both  being  equal  to  the  same  ratio? 
Can  this  be  possible  without  R  and  Q  coinciding? 

State  theorem  proved. 


182  PLANE  GEOMETRY  [VI,  §  215 

215.  Theorem  XII.  //  the  interior  angle  of  a  triangle  is 
bisected,  and  the  bisector  extended  until  it  cuts  the  opposite  side,  it 
divides  that  side  into  segments  proportional  to  the  other  two  sides. 


Given  the  triangle  ABC  with  the  angle  A  bisected  by  the 
line  ARy  forming  the  segments  BR  and  RC. 

To  prove  that  fc  =  £c. 

Analysis.  We  can  prove  this  theorem  by  forming  a  pencil 
of  lines  cut  by  a  pencil  of  parallels.  This  pencil  must  have 
BR  and  RC  for  segments  of  one  of  its  rays.  BA  can  be  a 
segment  on  another  ray.  This  suggests  immediately  that 
R  A  be  one  of  the  parallel  lines  and  that  we  must  draw  another 
line  parallel  to  RA  through  point  C.  Extending  BA  through 
A  until  it  meets  this  line  in  point  P,  we  have  a  proportion 
which  contains  BR,  RC,  and  BA,  AP.  These  are  all  of  the 
terms  in  the  proportion  that  we  are  trying  to  prove  excepting 
CA,  which  must  take  the  place  of  A  P.  So  that  if  our  pro- 
portion is  true  at  all,  we  must  be  able  to  prove  that  the  line 
CA  is  equal  to  the  line  AP. 

We  can  show  that  C  A  equals  AP  if  we  can  show  that  angle 
A  PC  equals  angle  PC  A.  We  can  show  angle  A  PC  equal  to 
angle  PC  A,  if  we  can  show  these  equal  respectively  to  the 
angles  BAR  and  RAC,  which  themselves  were  made  equal. 

Give  this  proof  complete. 

Exercise.  If  AB  =  AC,  what  fact  about  an  isosceles  tri- 
angle is  brought  out  by  this  theorem? 


VT,  §  216]  RATIO  AND  PROPORTION  183 

216.  Theorem  XIII.     //  the  exterior  angle  of  a  triangle  is 
bisected,  the  bisector  divides  the  opposite  side  externally  into 
segments  proportional  to  the  other  two  sides. 

Analyze  and  prove  on  the  same  plan  that  you  proved 
Theorem  XII. 

Notice  that  the  bisector  of  the  interior  angle  of  a  triangle 
divides  the  opposite  side  in  the  same  ratio  as  the  bisector  of 
the  exterior  angle  at  the  same  vertex.  That  is  the  external 
division  of  the  line  is  in  the  same  ratio  as  the  internal  division 
of  the  line. 

Definition.  When  a  line  is  divided  internally  and  ex- 
ternally in  the  same  ratio,  the  line  is  said  to  be  divided 
harmonically. 

217.  Exercises. 

1.  In  Theorem  I  show  that  the  rectangle  of  a  and  d  equals 
the  rectangle  of  b  and  c.     State  this  in  the  form  of  a  theorem. 

be          ad 

2.  Show  that  a  =  — ;  6  =  — .     (In  Theorem  I.) 

7-2 

3.  Construct  the  line  whose  length  is  equal  to  — . 

y 

4.  In  Theorem  I  if  a  =  r,  b  =  4  r  —  1,  c  =3,  and  d  =  12  r, 
find  the  value  of  r. 

5.  In  Theorem  I  if  a  =  m,  b  =2m  +  4,  c  =  6m  —  5,  d  = 
13  m  +  5,  find  the  value  of  m. 

6.  Two  parallel  lines  cut  a  pencil  of  two  rays,  cutting  the 
segments  a  and  b  on  one  ray,  and  c  and  d  on  the  other.     If 
segment  6  is  4  units  more  than  twice  segment  a,  segment  c  is 
5  units  less  than  6  times  segment  a,  and  segment  d  is  5  units 
more  than  13  times  segment  a,  find  the  lengths  of  segments 
a,  6,  c  and  d. 

Note.  In  solving  this  exercise  notice  that  it  leads  to  the 
algebraic  expressions  in  Exercise  5. 


184  PLANE  GEOMETRY  [VI,  §  217 

7.  In  Theorem  I  suppose  a  =  p,  b  =  \ p  —  1,  c  =  3  p,  and 
,~ i 

d  =  *-— — ,  write  a  problem  leading  to  these  expressions  and 

solve. 

8.  The  sides  of  a  triangle  are  7,  8,  and  12.     Find  the  length 
of  the  segments  into  which  the  bisector  of  each  angle  divides 
the  opposite  side.     Use  both  the  exterior  and  interior  angles. 
Construct  the  figure  and  check  results  by  measurements. 

9.  In  a  given  triangle  one  side  is  5|  units.     The  angle 
opposite  this  side  is  bisected  and  the  bisector  divides  this 
side  into  two  segments  such  that  if  4  units  be  added  to  the 
shorter  it  is  equal  to  one  of  the  sides  of  the  triangle,  and  if  4 
units  be  added  to  three  times  the  shorter  segment,  it  is  equal 
to  the  other  side.     Find  the  length  of  the  sides  of  the  triangle 
and  the  length  of  the  segments  into  which  the  first  side  is 
divided. 

10.  The  diagonals  of  a  trapezoid  divide  each  other  into 
parts  proportionate  to  the  bases  of  the  trapezoid. 

Suggestion.  Regard  the  point  of  intersection  of  the  diag- 
onals as  the  vertex  of  a  pencil  of  two  rays. 

11.  If  a  trapezoid  has  one  base  twice  as  long  as  the  other, 
into  what  ratio  will  the  diagonals  divide  each  other? 

12.  The  medians  of  a  triangle  meet  in  a  point  which  is  two- 
thirds  of  the  distance  from  any  vertex  to  the  mid-point  of  the 
opposite  side. 

Suggestion.  Show  that  this  is  an  application  of  Exercise  11 
by  noticing  that  the  line  joining  the  ends  of  two  medians  is 
parallel  to  the  third  side  of  the  triangle  and  half  as  long. 

13.  The  line  joining  the  mid-points  of  the  bases  of  a  trap- 
ezoid divides  each  of  the  other  sides  externally  into  parts 
proportional  to  the  bases. 

14.  A  line  segment  10  units  long  is  divided  internally  in 
the  ratio  2:3.     Determine  by  construction  an  external  point 
of  division  so  that  the  line  shall  be  divided  harmonically. 
Is  there  only  one  such  point? 


VI,  §  218J 


SIMILAR  FIGURES 


185 


PART  II— SIMILAR  FIGURES 
218.  Definition.     Draw  a  pencil  of  lines  0.     On  each  ray 


mark  a  point.  Call  these  points  AI,  BI, 
rays  mark  another  set  of  points  A2,  B2)  Cz 
OAi  OB1  OCi 


-  •  • .     On  the 

in  such  a  way 


that 


OA2      OB,      OC<> 


We  call  these  sets  of  points,  AI,  BI,  Ci  ....  and  A2,  B2,  Cz 
.  .  .  .  ,  similar  sets  of  points,  or  similar  systems  of  points. 

If  we  join  these  similar  systems  of  points  by  straight  lines, 
the  figures  formed  are  called  similar  polygons. 

Point  0  is  called  the  center  of  similitude.  The  ratio  of 
the  two  segments  on  any  ray  through  this  point  is  constant 
for  all  the  rays. 


Notation.   The  symbol  for  similar  is  ~ ,  an  s  lying  on  its  side. 

In  either  of  the  two  preceding  figures  we  have  A  AiBid 
«>  A  AzBzCz,  by  the  definition  of  similar  figures.  It  then 
follows  directly  from  Theorems  X  and  XI,  page  181,  that 
homologous  sides  of  these  triangles  are  proportional  and  that 
they  are  parallel,  and  therefore  that  the  triangles  are  mutually 


186 


PLANE  GEOMETRY 


[VI,  §  219 


equiangular.  (See  Corollaries  1  and  2  below.)  This  result 
is  often  used  as  the  starting  point  in  the  definition  of  similar 
triangles,  that  is,  two  triangles  are  defined  to  be  similar  when 
their  homologous  sides  are  proportional  and  their  homologous 
angles  are  equal. 

Corollary  1.     //  two  triangles  are  similar,  their  corresponding 
sides  i.re  proportional. 

c, 


Suggestion.     Prove  this  by  showing  that 

°L  =  2£  =JL  =  OA        c 
fli  "  OCi  ~  bi  ~  OAi  ~  d' 

Corollary  2.  //  two  triangles  are  similar,  they  are  mutually 
equiangular. 

Suggestion.     Use  Art.  214  and  Art.  62. 
Combining  corollaries  1  and  2,  we  have:  //  two  triangles 
are  similar,  they  are  mutually  equiangular  and  their  correspond- 
ing sides  are  proportional. 

219.  The  theorems  on  similar  triangles  are  very  closely  re- 
lated to  those  on  congruent  triangles.  In  fact,  as  will  be  seen, 
the  congruent  triangles  are  but  special  cases  of  similar  triangles. 
This  similarity  begins  in  the  treatment  of  the  two  subjects. 

In  beginning  the  discussion  of  congruent  triangles,  we  first 
defined  them  as  triangles  which  when  placed  one  upon  the  other 
could  be  made  to  coincide.  Then  we  proved  that  if  triangles 
had  certain  characteristics,  we  could  always,  if  we  chose,  place 
them  together  and  make  them  coincide.  Then,  if  we  could 
show  triangles  to  have  these  characteristics,  we  should  know 
them  to  be  congruent  without  placing  them  together. 


vi,  §  220]  SIMILAR  FIGURES  187 

So  in  the  treatment  of  similar  figures,  our  definition  assumes 
them  so  arranged  that  their  corresponding  points  lie  on  the 
rays  of  a  pencil,  whose  vertex  divides  these  rays  in  a  series  of 
equal  ratios.  The  figures  are  then  said  to  be  in  similar  per- 
spective. We  shall  prove  that  if  triangles  have  certain  char- 
acteristics, they  can  always  be  placed  in  similar  perspective, 
and  hence  it  will  not  be  necessary  to  so  place  them  in  order 
to  know  that  they  are  similar.  We  shall  know  them  to  be 
similar  by  these  characteristics. 

220.  Theorem  XTV.  Two  triangles  are  similar  if  they 
have  two  angles  of  the  one  equal  respectively  to  two  angles  of  the 
other. 

B, 


Given  the  triangle  ABC  and  A^d  with  the  angle  A  equal 
to  angle  AI,  and  angle  B  equal  to  angle  BI. 

To  prove  that  triangles  ABC  and  AiBiCi  are  similar. 

Analysis.     We  can  prove  that  A  ABC  and  AiB±C\  are 
similar  if  we  can  prove  that  their  sides  are  proportional,  that 

is,  prove  that  —  =  -=-  =  — .    We  can  show  that  —  -  — if,  when 
a*      bi      ci  al      ci 

we  place  the  two  triangles  together  with  the  equal  angles  B 
and  BI  coinciding,  we  can  prove  that  side  b  is  parallel  to  side 
61.  We  can  show  that  side  b  is  parallel  to  61  by  showing  that 
they  are  cut  by  the  transversal  AB,  and  that  the  correspond- 
ing angles  B  AC  and  BiAiCi  are  equal.  We  can  show  that 

T-  =  —  by  again  placing  the  triangles  together  with  the  equal 


188  PLANE  GEOMETRY  [VI,  §  220 

angles  A  and  A!  coinciding  and  showing  that  the  sides  a  and 
ai  are  parallel. 

Proof.     Place  &  ABC  and  A^d  together  so  that  Z  B 
will  coincide  with  Z  BI.     (Why  will  they  coincide?) 

Then  point  AI  will  fall  on  side  c  or  on  side  c  produced,  and 
point,  Ci  will  fall  on  side  a  or  on  side  a  produced. 

Z  BAG  =  ZAAiCi.  Why? 

6  ||  61.  Why? 

H-  why? 

In  like  manner,  placing  the  triangles  together  with  the 
equal  angles  A  and  A\  coinciding,  prove  that  —  =  -  . 


A  ABC  ~  A  AiBiCi.  Why? 

State  the  theorem  proved. 

Corollary  1.  Two  triangles  are  similar  if  they  have  their  sides 
parallel  each  to  each,  or  their  sides  perpendicular  each  to  each. 

Corollary  2.     Two  equilateral  triangles  are  similar. 

Corollary  3.  Two  right  triangles  are  similar  if  they  have  an 
acute  angle  of  the  one  equal  to  an  acute  angle  of  the  other. 

Exercises. 

1.  Construct  a  triangle  similar  to  a  given  triangle,  using 
the  above  theorem. 

2.  Construct  a  triangle  similar  to  a  given  triangle,  using 
each  of  the  above  corollaries. 

3.  The  perpendicular  from  the  vertex  of  the  right  angle  of 
a  right  triangle  to  the  hypotenuse  forms  two  right  triangles 
similar  to  the  first  triangle. 

4.  The  sides  of  a  triangle  are  each  4  times  as  long  as  the  sides 
of  another  triangle.     What  can  you  say  about  their  angles? 

5.  If  A  ABC  ™  A  AiBiCi,  and  if  2  AB  =  3  A^,  what  is 
the  ratio  of  the  perimeters?     Construct  two  such  triangles. 


VI,  §  221]  SIMILAR  FIGURES  189 

6.  Construct  a  triangle  with  sides  3,  5,  6  units  long  re- 
spectively, and  prolong  each  side  in  both  directions.  By 
drawing  in  three  more  lines  construct  three  triangles  with 
sides  twice  as  long  as  those  of  the  first  triangle  and  similar 
to  it  and  to  each  other. 

221.  Theorem  XV.     The  corresponding  altitudes  of  similar 
triangles  are  proportional  to  their  corresponding  sides. 

Suggestion.    Apply  Art.  220,  Cor.  3. 

222.  Theorem  XVI.     //  two  triangles  have  an  angle  of  the 
one  equal  to  an  angle  of  the  other,  and  the  including  sides  pro- 
portional, the  triangles  are  similar. 


Given  the  triangles  ABC  and  AiB^i  with  angle  A  equal 
to  angle  AI  and  the  including  sides  of  such  length  that 


To  prove  that  triangles  ABC  and  AiBiCi  are  similar. 

Analysis.  We  can  prove  this  if  we  can  prove  two 
angles  of  one  equal  respectively  to  two  angles  of  the  other. 
Since  Z  A  equals  Z  AI,  we  have  but  to  prove  that  Z  B  equals 
Z  BI.  By  placing  the  two  triangles  together  with  equal 
angles  A  and  AI  coinciding,  we  can  prove  that  A  B  and  #1  are 
equal  if  we  can  prove  that  side  a  is  parallel  to  side  01.  This 
we  can  prove  by  showing  that  side  ai  divides  the  sides  b  and  c 
proportionally. 

Proof.     Give  a  complete  proof  of  this  theorem. 

Exercise.  Construct  a  triangle  similar  to  a  given  triangle, 
using  the  above  theorem. 


190 


PLANE  GEOMETRY 


[VI,  §  223 


223.  Theorem  XVII.     If  two  triangles  have  their  sides  pro- 
portional they  are  similar. 


Given  the  triangles  ABC  and  AiB^Ci  with  their  sides  of 

such  length  that  —  =  r  =  —  . 
ai      61      ci 

To  prove  that  triangles  ABC  and  AiBiCi  are  similar. 

Analysis.  Since  we  have  no  angles  given  in  this  proposi- 
tion, we  shall  not  be  able  to  place  the  triangles  together,  and 
know  that  their  sides  will  take  the  same  direction.  We  shall 
make  a  pencil  with  vertex  A.  On  ray  AC  we  lay  off  AK 
equal  to  61.  On  ray  AB  we  lay  off  AH  equal  to  ci.  Join  H 
and  K.  We  can  now  prove  our  proposition  if  we  can  prove 
A  AH  K  and  ABC  similar,  and  then  prove  that  A  AH  K  is 
congruent  to  A  AiB\C\. 


Proof. 
In 

Since 
But 


HK  ||  BC.     (See  Theorem  XI.)  Why? 
A  AUK  ~  A  ABC.  Why? 


AK  =61, 

AH  =  ci,  by  construction. 

AAHK™  A  ABC, 

c_         a 
*~#7r 


ff  #  =  a,. 


Why? 
Why? 

Why? 
Why? 


VI,  §  224]  SIMILAR  FIGURES  191 

A  AHK  ^  A  AiBA. 
A  AitfiCi  ~  A  ABC.  Why? 

224.  Exercises. 

1.  Construct  a  triangle  similar  to  a  given  triangle  but  with 
half  the  perimeter. 

2.  Construct  a  triangle  and  draw  the  lines  joining  the  mid- 
points of  its  sides.     Show  that  the  four  triangles  so  formed 
are  similar  to  the  given  triangle  and  congruent  to  each  other. 

3.  Construct  a  triangle  similar  to  a  given  triangle  and  hav- 
ing its  perimeter  equal  to  a  given  straight  line. 

Suggestion.  In  the  figure  of  Art.  87,  make  AB  equal  to 
the  given  perimeter;  then  lay  off  on  AX  segments  equal  to 
the  sides  of  the  given  triangle  instead  of  equal  segments. 

State  theorems  on  congruent  triangles  that  correspond  to 
theorems  on  similar  triangles,  and  explain  how  the  former 
are  special  cases  of  the  latter. 

225.  Theorem  XVIII.     //  two  polygons  are  similar  they 
can  be  divided  into  the  same  number  of  similar  triangles,  simi- 
larly placed. 


Suggestion.  Since  the  polygons  are  similar  they  can  be 
placed  in  similar  perspective.  Then  the  triangles  are  similar 
by  definition. 

Corollary  1.  //  two  polygons  are  similar,  they  are  mutually 
equiangular,  and  their  corresponding  sides  are  proportional. 

Corollary  2.  The  perimeters  of  similar  polygons  have  the 
same  ratio  as  any  two  corresponding  sides. 

Suggestion.  State  the  second  part  of  Corollary  1  in  the 
form  of  a  series  of  equal  ratios  and  apply  Theorem  VII , 


192  PLANE  GEOMETRY  [VI,  §  226 

226.  Theorem  XIX.  //  two  polygons  are  mutually  equi- 
angular, and  have  their  corresponding  sides  proportional,  they 
are  similar. 


Given  the  two  polygons  A  BCD  ......  and  AiBiCiDi  ...... 

AB       BC       CD       DE 

with  sides  of  such  length  that  T-^-  =  -^-^  =  7^77  =  =-=  = 

AI&I       ^>iCi       CI/A       L)\rji 

......  ,  and  with  angles  such  that  Z  A  =  Z  A1}  Z  B  =  Z  Bi, 

Z  C  =  Z  Ci,  ......  . 

To  prove   that  polygon  ABCD    ......   is  similar  to  polygon 


Analysis.  We  can  show  these  two  polygons  similar  if  we 
can  show  that  they  can  be  so  placed  that  the  joins  of  their 
corresponding  points  will  form  a  pencil  of  lines.  We  start  by 
placing  the  side  AB  parallel  to  side  AiBi,  then  the  remaining 
corresponding  sides  will  fall  parallel  to  one  another  respect- 
ively. The  joins  of  A,  A\  and  B,  BI  will  meet  in  a  point. 
Why?  Let  this  point  be  V.  Now  if  we  can  prove  that  join 
of  C,  Ci  will  meet  BBi  in  the  same  point,  we  will  have  proved 
our  proposition.  Let  V  be  the  point  in  which  CC\  meets 
BBi.  We  can  prove  that  V  and  V  coincide  if  we  can  prove 

A,    .  BV       BV  ,  .  BC 

that  -5-Y,  =  -H^FT/  by  proving  them  each  equal  to  -5-^-  . 

D\  V  £>lV  .  jDlOl 

Give  this  proof  complete. 

In  the  figure  above  it  is  understood  of  course  that  the  lines 
there  shown  represent  only  portions  of  the  entire  polygons, 
which  may  have  any  number  of  sides. 


VI,  §  227] 


SIMILAR  FIGURES 


193 


227.  Exercises. 

1.  The  sides  of  a  triangle  are  5,  7,  and  10.     Construct  a 
similar  triangle  whose  side  corresponding  to  side  5  is  11, 

2.  On  cross  section  paper  draw  a  map  of  a  state,  with  ir- 
regular boundary  lines  straightened  out,  keeping  the  area 
about  the  same. 

3.  Take  measurements  of  the  ground  outline  of  your  school 
building.     Reproduce  it  in  correct  proportions  on  cross  sec- 
tion paper. 

4.  Suppose  that  in  measuring  from  one  point  to  another, 
an  obstacle  is  in  your  way  so  that  you 

cannot  go  directly.     Show  that  by  taking 
measurements  as  in  the  figure,  and  draw- 
ing to  a  suitable  scale,  you  can  find  the 
length  desired  by  measurement  of  the 
figure. 

Make  such  measurements. 

5.  Show  how,  by  measuring  the  sides 
of  a   triangular   piece  of  ground,  you 
could  determine  the  angles  by  construc- 
tion and  measurement. 

6.  In  the  adjacent  figure  line  BC  is 
drawn  so  that  Z  ABC  =  /.  ADE.    Find 
the  width  of  the  river.      Check  by 
drawing  a  figure  to  scale. 

7.  In  the  adjacent  figure  prove  that  A 
BC  is  parallel  to  DE.     Calculate  the 
length  of  DE.     Check  by  drawing  the 
figure  to  scale. 

8.  In  the  adjoining  figure  prove 
that  A  C  is  parallel  to  DE.    Calculate 
the  length  of  DE.    Check  by  draw-  A 
ing  the  figure  to  scale. 


194  PLANE  GEOMETRY  [VI,  §  227 

9.  Let  the  student  go  out  and  take  measurements,  for  the 
purpose  of  computing  a  distance  that  cannot  be  measured. 

10.  A  boy  found  that  by  holding  a  pencil  and  a  foot-rule 
at  different  distances  from  his  eye  but  parallel  to  each  other, 
he  could  make  the  length  of  the  pencil  just  cover  the  length 
of  the  ruler.     In  this  position  the  distance  from  his  eye  to  one 
end  of  the  pencil  was  4  in.  more  than  the  length  of  the  pencil, 
while  the  distance  to  the  corresponding  end  of  the  ruler  was 
2  in.  more  than  twice  the  length  of  the  pencil.     Find  the 
length  of  the  pencil. 

11.  A  triangular  flower  bed  is  surrounded  by  a  uniform 
walk;  the  entire  plot  is  fenced  in.     The  length  of  the  fence 
along  the  shortest  side  is  12  ft.     The  length  of  the  shortest 
side  of  the  bed  is  one  foot  less  than  half  the  next  side  and  the 
fencing  along  that  side  is  9  ft.  longer  than  the  side.     If  the 
entire  length  of  fencing  is  61  f  ft.  find  the  length  of  each  side 
of  the  flower  bed. 

12.  Write  a  problem  involving  similar  triangles  which  will 
lead  to  a  quadratic  equation  and  solve.     (See  Exercise  11.) 

13.  Write  a  problem  involving  Theorem  XII,  which  will 
lead  to  a  quadratic  equation.     Solve. 

14.  Write  a  problem  involving  Theorem  X  which  will  lead 
to  a  quadratic  equation.     Solve. 

16.  The  altitudes  of  a  triangle  divide  each  other  propor- 
tionally. In  the  proof  use  triangles  containing  both  acute 
and  obtuse  angles. 

16.  Draw  a  circle,  two  intersecting  chords  in  the  circle, 
and  chords  joining  the  ends  of  the  intersecting  chords.    Name 
and  prove  similar  triangles  formed.     See  Art.  228,  p.  195. 

17.  Draw  a  circle  with  any  radius,  say  2  inches,  and  in  it 
draw  a  chord  3  inches  long.     Through  a  point  1  inch  from 
the  end  of  this  chord  draw  another  chord  which  shall  be 
divided  at  this  point  in  the  ratio  3:1. 

Suggestion.  First  calculate  the  segments  of  the  second 
chord  by  use  of  Exercise  16,  then  construct  them. 


VI,  §  228]  SIMILAR  FIGURES  195 

228.  Theorem  XX.     //  two  intersecting  chords  are  drawn  in 
a  circle,  the  product  of  the  segments  of  one  chord  is  equal  to  the 
product  of  the  segments  of  the  other. 

Suggestion.  Draw  chords  to  join  the 
ends  of  the  intersecting  chords  and  prove 
triangles  similar  as  in  Exercise  16.  Form 
a  proportion,  and  clear  the  equation  of 
fractions. 

This  proposition  is  frequently  stated: 

//  a  pencil  of  lines  is  cut  by  a  circumference,  the  product  of 
the  two  segments  cut  from  the  vertex  of  the  pencil  is  constant  no 

matter  which  line  is  taken.  p 

^f 
Prove   that   this    statement  is  true 

when  the  vertex  of  the  pencil  is  with- 
out the  circumference.  D[ 

Prove  triangles  DBF  and  AFC  simi- 
lar. 

By  turning  one  of  the  rays  gradually  about  the  vertex  of 
your  pencil  show  that: 

Corollary.  The  tangent  from  the  vertex  of  a 
pencil  of  lines  is  a  mean  proportional  between 
the  segments  of  any  other  ray  of  the  pencil. 

229.  Exercises. 

1.  The  square  of  half  the  shortest  chord  that  can  be  drawn 
through  a  point  within  a  circle  is  equal  to  the  product  of  the 
segments  of  the  diameter  through  the  point. 

This  may  be  stated: 

One-half  of  the  shortest  chord  drawn  through  a  point 
within  a  circle  is  a  mean  proportional  between  the  segments 
of  the  longest  chord  through  that  point. 

Another  statement  is: 

The  perpendicular,  drawn  from  the  vertex  of  the  right 
angle  of  a  right  triangle  to  the  hypotenuse,  is  a  mean  pro- 
portional between  the  segments  of  the  hypotenuse. 


196  PLANE  GEOMETRY  [VI,  §  230 

2.  Explain  how  the  above  exercise  may  be  used  to  construct 
a  mean  proportional  between  two  given  lines. 

3.  How  far  is  a  point  from  a  circular  lake  7  miles  in  diam- 
eter, if  the  distance  from  the  point  to  the  shore  measured 
along  the  tangent  line  is  6  miles  less  than  twice  the  distance 
measured  along  a  line  which,  if  continued,  would  pass  through 
the  center  of  the  lake. 

Solve  Exercise  3  without  using  the  Pythagorean  theorem. 

4.  Write  a  problem  involving  Theorem  XX,  which  will 
lead  to  a  quadratic  equation.     Solve. 

5.  If  two  circles  are  tangent  and  three  secants  are  drawn 
through  the  point  of  tangency,  prove  that  the  chords  join- 
ing the  ends  of  these  secants  form  similar  triangles. 

6.  At  the  ends  of  a  diameter  of  a  circle,  with  center  C  and 
radius  r,  erect  perpendiculars  to  it.     Draw  a  line  tangent  at 
any  point  P  and  cutting  the  Js  at  A  and  B.     Prove  A  ABC 
is  right-angled  and  that  AP  -  PB  =  r2. 

7.  Turn  to  the  figure  of  Ex.  24,  Art.  168,  and  name  all  of 
the  similar  triangles  that  you  can  find,  and  all  of  the  pro- 
portions, giving  reasons. 

230.  Theorem  XXI.     //  a  perpendicular  is  dropped  from  the 
vertex  of  the  right  angle  of  a  right  triangle  to  the  hypotenuse: 

(a)  Two  triangles  will  be  formed  which  are  similar  to  the 
original  triangle  and  similar  to  each  other; 

(b)  Either  side  is  a  mean  proportional  between  the  hypotenuse 
and  the  adjacent  segment; 

(c)  The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 


A 

Analyze,  and  prove. 


vi,  §  231]  SIMILAR  FIGURES  197 

Problem  V.     To  construct  a  mean  proportional  between  two 
given  lines. 


Given  the  lines  m  and  n. 

To  construct  a  line  which  is  a  mean  proportional  between  m 
and  n. 

Analysis.  Suppose  the  figure  drawn.  We  see  that  we 
shall  have  PC  a  mean  proportional  between  m  and  n,  if  we 
can  have  angle  C  a  right  angle  and  AB  the  hypotenuse  of  a 
right  triangle,  AB  being  the  sum  of  m  and  n,  and  P  being  the 
point  which  separates  these  two  segments.  That  is  we  are 
required  to  locate  a  point  C,  from  the  conditions  given,  first 
that  C  must  lie  on  the  perpendicular  drawn  to  A  B  at  point  P, 
second  that  C  must  be  the  vertex  of  a  right  angle  of  a  triangle 
of  which  A  B  is  the  hypotenuse. 

The  locus  from  the  first  condition  is  the  perpendicular. 
The  locus  to  fulfil  the  second  condition  is  the  circumference 
on  A  B  as  a  diameter.  The  intersection  of  these  two  lines 
is  the  required  point  C. 

Make  this  construction  and  prove. 

Corollary.     To  construct  a  square  equal  to  a  given  rectangle. 

231.  Exercises. 

1.  Construct  a  square  which  equals  half  a  given  rectangle. 

2.  Construct  a  square  which  equals  twice  the  square  con- 
structed in  Exercise  1.     How  does  this  square  compare  with 
the  given  rectangle  of  Exercise  1? 

3.  Construct  a  square  which  is  one-fifth  of  a  given  square. 

4.  The  diagonal  of  one  square  is  three  times  the  diagonal 
of  a  second  square.     Find  the  ratio  of  their  areas. 


198  f  PLANE  GEOMETRY  [VI,  §  231 

5.  The  area  of  a  square  is  10  square  units.     Find  the 
length  of  the  line  drawn  from  one  corner  to  the  middle  of 
one  of  the  opposite  sides. 

6.  A  line  CD  of  varying  length,  perpendicular  to  A  B,  moves 
with  its  end  C  along  AB  so  that  'CD 2  =  AC-  CB.     Find  the 
locus  of  point  D. 

7.  In  a  right  triangle  ABC,  Z  A   =  45°.     If  Z  A  is  bi- 
sected, and  the  bisector  cuts  side  EC  at  D,  find  the  ratio  of 
BD  to  DC. 

8.  If  a  circle  is  inscribed  in  a  right  triangle,  the  sum  of  diam- 
eter and  hypotenuse  equals  the  sum  of  the  other  two  sides. 

9.  In  right  triangle  ABC,  let  AB  =  c,  AC  =  b,  CB  =  a; 
CG  is  _L  AB,  and  A E,  BD,  CF  are  medians. 

Show  that  CG  =  — ;  AG  =  -;  GB  =  -; 
c  c  c 


=  J  \/462  +  a2; 
BD  =  |  V4a2+62;  CF  =  Jc; 

radius  of  circumscribed  O  =  Jc; 

a  _L  5  _  c  • 

radius  of  inscribed  O  =  — . 

& 

10.  Draw  an  accurate  figure  like  that  in  Ex- 
ercise 9,  taking  a  =  2  in.,  and  b  =  4  in.     Calculate  the  lengths 
of  all  the  lines,  and  check  by  measurement. 

11.  If  r  and  rr  are  the  radii  of  two  concentric  circles,  re- 
spectively, find  the  length  of  a  chord  of  the  outer  circle  that 
is  tangent  to  the  inner  circle.     First  use  the  Pythagorean 
theorem  in  the  solution,  then  solve  again  by  use  of  Theorem 
XX,  page  195.     To  apply  this  theorem  draw  one  of  the 
chords  mentioned,  and  through  the  point  where  it  touches 
the  inner  circle  draw  a  diameter  of  the  outer  circle  per- 
pendicular to  the  chord. 

In  the  following  find  the  mean  proportional  by  drawing  and 
by  Algebra.  Compare  results,  by  counting  if  rational,  by 
constructing  by  the  Pythagorean  theorem  if  irrational. 


vi,  §  232]  SIMILAR  FIGURES  199 

12.  9  and  225.        13.  8  and  72.  14.  34  and  136. 

15.  6  and  3.  16.  V 5  and  V 15.          17.  V^IandVH 

Without  drawing  find  the  mean  proportional  between 
18.  a  and  b.  19.  a2and&2. 

a2  -  62  ,  a3  +  63 

20-  ~t .    ,   TO  and r-. 

a2  —  a&  +  62          a  —  6 

21.  A  railroad  forms  a  chord  across  a  circular  lake  whose 
diameter  is  d.     To  go  to  the  half-way  station  one  must  go 
along  the  railroad  a  feet  more  than  n  times  as  far  as  to  go  by 
water  from  a  point  on  the  shore  directly  opposite  the  station. 
How  far  is  it  to  the  station  by  each  route. 

22.  Write  Exercise  21  with  special  values  for  the  general 
numbers,  and  use  the  answers  of  that  exercise  as  formulae  to 
solve  your  exercise. 

23.  Write  a  general  problem  involving  Theorem  XXI,  lead- 
ing to  a  quadratic  equation.     Solve. 

24.  Rewrite  Exercise  23,  giving  special  values  to  the  gen- 
eral numbers  used,  and  solve  by  using  the  answers  to  Exer- 
cise 23  as  formulas. 

232.  Up  to  this  time  we  have  been  obtaining  the  lengths 
of  lines  representing  surd  numbers  by  means  of  the  Pythag- 
orean theorem — but,  as  has  been  remarked,  this  is  not  the 
practical  method.  Approximate  values  as  close  as  desired 
are  used  in  practice.  For  the  purpose  of  close  measure- 
ment various  instruments  have  been  devised.  A  very  simple 
one  which  you  can  make  and  use  in  the  future  is  made  as 
follows:  • 

Line  in  a  square  of  10  units  on  your  cross- 
section  paper.  Numbering  the  division 
points  each  way  0, 1,  2,  .  . .  . ,  join  the  tenth  *| 
point  at  the  top  to  the  ninth  point  at  the 
bottom,  and  so  on.  Rule  your  own  cross- 
section  paper  if  none  is  at  hand. 


200  PLANE  GEOMETRY  pro.  §  233 

Regard  the  length  of  your  square  as  1  unit,  and  explain  how 
your  diagonal  scale  may  be  used  to  measure  any  number  of 
tenths  or  hundredths  of  a  unit.  Give  reason. 

Make  a  diagonal  scale  by  which  any  number  of  tenths  or 
hundredths  of  an  inch  may  be  measured. 

Explain  how  your  compass  may  be  used  to  take  a  required 
length  stated  in  units,  tenths,  and  hundredths,  from  the 
diagonal  scale  and  lay  it  off  on  a  straight  line.  If  greater 
accuracy  is  required,  dividers  instead  of  the  compass  should 
be  used. 

PART  III— TRIGONOMETRIC  FUNCTIONS 

233.  In  the  history  of  mathematics  we  read  that  Thales 
measured  the  heights  of  the  Pyramids  in  Egypt.  He  did  it 
in  this  way:  At  the  time  of  day  when  he  observed  that  the 
length  of  the  shadow  of  a  stick  set  perpendicularly  to  the 
ground  was  as  long  as  the  stick,  he  measured  the  length  of 
the  shadow  of  the  pyramid.  Did  he  work  on  correct  mathe- 
matical principles? 

Using  the  same  plan  that  Thales  used,  take  measurements 
and  find  the  height  of  a  smoke  stack  or  some  building,  a  tele- 
phone pole  or  electric  light  pole. 

It  will  not  be  necessary  for  you  to  wait  until  the  shadow  is 
the  same  length  as  the  stick  that  you  use.  You  may  take 
the  measurement  at  any  convenient  time.  Why?  How  does 
Thales'  method  show  that  he  was  limited  in  his  knowledge  of 
the  subject  of  proportion? 

The  plan  that  Thales  used  required  that  the  sun  be  shining, 
so  as  to  produce  shadows.  This  plan  would  not  work  in  a 
forest. 

A  man  measured  the  heights  of  trees  with  a  home-made  in- 
strument of  the  following  sort.  He  had  two  strips  of  wood 
hinged  together  at  the  ends.  Running  the  length  of  one  of 
these  strips  was  a  groove  in  which  another  strip  worked 


VI,  §  234] 


TRIGONOMETRIC  FUNCTIONS 


201 


smoothly.  When  he  was  ready  to  use  this  instrument,  he 
would  push  one  of  the  hinged  strips,  which  had  a  sharp  end, 
into  the  ground  and  turn  the  N 
other  strip  and  sight  along  it 
until  it  pointed  to  the  top  of 
the  tree.  He  would  then  allow 
the  strip  in  the  groove  to  slide 
down  until  it  touched  the  L  Q  P 

ground,  which  point  he  marked.  With  a  steel  tape  he 
measured  from  this  point  to  the  foot  of  the  stake  and  on  to 
the  foot  of  the  tree.  From  the  data  he  computed  the  height 
of  the  tree. 

On  what  principle  did  he  work? 

Exercise.  If  RQ  is  5  feet,  PQ  is  10  feet,  PL  is  225  feet, 
how  high  is  the  tree? 

Find  the  height  of  the  tree  if  PQ  =  a,  PL  =  b,  and  QR  =  c. 

This  method  depends  on  the  use  of  similar  right  triangles, 
and  is  much  used  in  measurements  of  heights  and  distance. 

234.  In  each  of  the  methods  described  above  it  was  neces- 
sary to  use  two  triangles  in  the  solution.  We  shall  now 
investigate  methods  by 
which  we  can  compute 
heights  and  distances  from 
one  triangle. 

Draw  two  lines  of  refer- 
ence at  right  angles  to  one 
another.  These  divide  the 
plane  of  your  paper  into 
four  quarters  or  quadrants. 
Art.  184.  Select  point  P 
any  point  in  the  plane. 
Four  cases  arise  according  to  the  quadrant  in  which  you 
select  point  P.  These  are  shown  in  the  figure. 


202 


PLANE  GEOMETRY 


[VI,  §  234 


With  reference  to  any  one  of  these  points,  three  lines  are 
to  be  noted,  namely,  the  abscissa,  the  ordinate,  and  the 
distance  from  0.  For  the  point  P  in  general  these  are  indi- 
cated by  the  letters  x,  y,  r  respectively.  For  point  PI  we  use 
the  letters  xi}  yi,  ri}  and  so  on  for  other  points. 

The  student  should  carefully  notice  the  signs  of  these 
quantities.  Thus  in  the  above  figure  Xi  stands  for  a  positive 
number,  xz  for  a  negative  number.  Also  y\  and  i/2  stand  for 
positive  numbers,  and  ys  and  2/4  stand  for  negative  numbers. 
The  distances  r\t  r2)  rS)  r4  are  all  considered  positive. 

Among  these  three  quantities  there  are  six  possible  ratios. 
Thus  for  x}  y}  r  we  have  the  ratios: 

y        x       y_       r_       r        x 
r'      r '      x'      y'      x"*      y' 

The  last  three  ratios  are  merely  the  reciprocals  of  the  first 
three. 


From  the  above  figure,  measure  the  lengths  of  x,  y,  r  for 
each  of  the  points,  give  each  length  its  proper  sign  and  then 
form  the  ratios;  arrange  your  work  in  a  table,  as  in  Art.  235 
below. 


VI,  §  235] 


TRIGONOMETRIC  FUNCTIONS 


203 


Thus  for  the  point  P2,  we  find 

x  =  -  6,          y  =  +  8,          r  =  10. 

Hence,     -  =  +  0.80,  -  0.60,     ^  =  -  1.33,  and  so  on. 

The  question  now  arises,  on  what  do  these  ratios  depend? 

Draw  a  careful  figure  showing  a  point  P  in  the  first  quad- 
rant, as  PI  in  the  figure  above. 

Measure  x,  y,  r  and  calculate  the  ratio  to  the  hundredths 
place. 

Extend  line  OP\  and  on  it  mark  a  new  point  P'.  Call  the 
new  values  x',  y',  r'.  Measure  and  calculate  as  before. 

Are  the  ratios  the  same  as  before?  Give  a  geometric  proof 
that  they  must  be  so. 

235.  Then  we  discover  this  truth, 
that  so  long  as  the  angle  XOP  remains 
the  same,  the  ratios  remain  the  same, 
even  though  the  length  of  x,  y}  r  are 
changed. 

Allow  r  to  remain  the  same  and 
turn  it  about  the  origin  until  it  makes  °. 
an  angle  of  10°  with  the  axis  of  the 
abscissa.  This  fixes  the  length  of  x  and  y.  Take  measure- 
ments and  find  the  six  ratios  to  hundredths  place.  Place  in 
the  form  of  a  table,  thus: 


Angle 

x 

y 

r 

y. 

r 

x 
r 

y. 

X 

r 

y 

r 
x 

x 

y 

Turn  r  about  the  origin  until  it  makes  an  angle  of  80°  with 
the  axis  of  the  abscissa.  Take  measurements  and  find  the 
ratios  as  before.  Does  a  change  in  the  angle  cause  a  change 
in  the  ratios? 


204  PLANE  GEOMETRY  [VI,  §  235 

Repeat  the  above  for  the  angles  20°;  70°;  30°;  60°;  40°;  50°; 
45°. 

Do  you  notice  any  relation  between  any  of  these  ratios? 
Prove  that  this  relation  should  exist. 

In  this  work  what  is  the  independent  variable?  What  are 
the  dependent  variables?  May  we  then  say  that  these  ratios 
are  functions  of  the  angle? 

Mathematicians  have  given  names  to  these  ratios. 

—  is  called  the  sine  of  the  angle. 

—  is  called  the  cosine  of  the  angle. 

—  is  called  the  tangent  of  the  angle. 
x 

—  is  called  the  cotangent  of  the  angle. 
y 

—  is  called  the  secant  of  the  angle. 

gp 

T 

—  is  called  the  cosecant  of  the  angle. 

If  we  let  0  (Greek  letter  Phi)  be  the  number  of  degrees  in 
the  angle,  these  functions  are  usually  written: 

qj  />» 

sin<£=— .  cot<£=— , 

r'  y' 

x  r 

— ,  sec<£=  — , 

T  X 

— .  csc<£=— . 

x'  y 

Note.  The  student  should  have  a  care,  in  speaking  or 
writing  these  functions  of  the  angle,  not  to  drop  the  angle. 
There  is  no  such  a  thing  as  a  sine  in  itself.  It  is  always  the 
sine  of  an  angle.  So  with  the  other  functions. 

Compare  your  table  with  the  one  found  on  following 
page. 


VI,  §  236] 


TRIGONOMETRIC  FUNCTIONS 


205 


236. 


Table  of  Trigonometric  Functions. 


sin  </> 

COS  <f> 

tan  <£ 

cot  <f> 

sec  </> 

cosec  </> 

0° 

0.000 

1.000 

0.000 

00 

1.000 

00 

90° 

1 

0.017 

0.999 

0.017 

57.290 

1.000 

57.299 

89 

2 

0.035 

0.999 

0.035 

28.636 

1.001 

28  .  654 

88 

3 

0.052 

0.999 

0.052 

19.081 

1.001 

19.107 

87 

4 

0.070 

0.998 

0.070 

14.301 

1.002 

14.335 

86 

5° 

0.087 

0.996 

0.087 

11.430 

1.004 

11.474 

85° 

6 

0.104 

0.995 

0.105 

9.514 

1.006 

9.567 

84 

7 

0.122 

0.993 

0.123 

8.144 

1.008 

8.206 

83 

8 

0.139 

0.990 

0.140 

7.115 

1.010 

7.185 

82 

9 

0.156 

0.988 

0.158 

6.314 

1.012 

6.393 

81 

10° 

0.174 

0.985 

0.176 

5.671 

1.015 

5.759 

80° 

11 

0.191 

0.982 

0.194 

5.145 

1.019 

5.241 

79 

12 

0.208 

0.978 

0.213 

4.705 

1.022 

4.810 

78 

13 

0.225 

0.974 

0.231 

4.332 

1.026 

4.445 

77 

14 

0.242 

0.970 

0.249 

4.011 

1.031 

4.134 

76 

15° 

0.259 

0.966 

0.268 

3.732 

1.035 

3.864 

75° 

16 

0.276 

0.961 

0.287 

3.487 

1.040 

3.628 

74 

17 

0.292 

0.956 

0.306 

3.271 

1.046 

3.420 

73 

18 

0.309 

0.951 

0.325 

3.078 

1.051 

3.236 

72 

19 

0.326 

0.946 

0.344 

2.904 

1.058 

3.072 

71 

20° 

0.342 

0.940 

0.364 

2.748 

1.064 

2.924 

70° 

21 

0.358 

0.934 

0.384 

2.605 

1.071 

2.790 

69 

22 

0.375 

0.927 

0.404 

2.475 

1.079 

2.670 

68 

23 

0.391 

0.921 

0.424 

2.356 

1.086 

2.559 

67 

24 

0.407 

0.914 

0.445 

2.246 

1.095 

2.459 

66 

25° 

0.423 

0.906 

0.466 

2.144 

1.103 

2.366 

65° 

26 

0.438 

0.899 

0.488 

2.050 

1.113 

2.281 

64 

27 

0.454 

0.891 

0.510 

1.963 

1.122 

2.203 

63 

28 

0.469 

0.883 

0.532 

1.881 

1.133 

2.130 

62 

29 

0.485 

0.875 

0.554 

1.804 

1.143 

2.063 

61 

30° 

0.500 

0.866 

0.577 

1.732 

1.155 

2.000 

60° 

31 

0.515 

0.857 

0.601 

1.664 

1.167 

.942 

59 

32 

0.530 

0.848 

0.625 

1.600 

1.179 

.887 

58 

33 

0.545 

0.839 

0.649 

1.540 

1.192 

.836 

57 

34 

0.559 

0.829 

0.675 

1.483 

1.206 

.788 

56 

35° 

0.574 

0.819 

0.700 

1.428 

1.221 

.743 

55° 

36 

0.588 

0.809 

0.727 

1.376 

1.236 

.701 

54 

37 

0.602 

0.799 

0.754 

1.327 

1.252 

.662 

53 

38 

0.616 

0.788 

0.781 

1.280 

1.269 

.624 

52 

39 

0.629 

0.777 

0.810 

1.235 

1.287 

.589 

51 

40° 

0  .  643 

0.766 

0.839 

1.192 

1.305 

.556 

50° 

41 

0.656 

0.755 

0.869 

1.150 

1.325 

.524 

49 

42 

0.669 

0.743 

0.900 

1.111 

1.346 

.494 

48 

43 

0.682 

0.731 

0.933 

1.072 

1.367 

.466 

47 

44 

0.695 

0.719 

0.966 

1.036 

1.390 

.440 

46 

45° 

0.707 

0.707 

1.000 

1.000 

1.414 

1.414 

45° 

cos  4> 

sin  '/> 

cot  <J> 

tan  </> 

cosec  <f> 

sec(/> 

206  PLANE  GEOMETRY  [vi,  §  237 

237.  Examining  the  table  we  find  that  it  gives  functions 
only  for  angles  measured  in  degrees.  We  shall  now  study 
its  use  in  finding  functions  of  angles  measured  in  degrees  and 
minutes. 

Example  1.     To  find  the  sin  35°  13 '. 

Looking  at  the  table  you  find  two  columns  marked  degrees, 
the  first  and  last  columns.  Beginning  at  the  top  the  first 
column  contains  all  of  the  angle  from  0  to  45  degrees.  Begin- 
ning at  the  bottom  the  last  contains  all  of  the  angle  from  45 
degrees  to  90  degrees  reading  upward.  From  the  work  called 
for  in  Art.  235  explain  why  the  table  may  be  made  in  this 
way,  the  names  at  the  top  belonging  to  the  angles  of  the  first 
column,  and  the  names  at  the  bottom  belonging  to  the  angles 
of  the  last  column. 

Running  down  the  first  column  you  find  35°.  Looking  in 
the  second  column  (marked  sin  <£  at  the  top)  in  the  same  line 

we  find 

sin  35°  =  .574. 

Also  sin  36°  =  .588. 

Our  angle  lies  between  these  two  angles,  so  we  argue  that  its 
sine  must  lie  between  .574  and  .588.  We  also  notice  that  as 
the  angle  increases  1°,  which  is  60',  the  sine  of  the  angle  in- 
creases .014.  Therefore  for  an  increase  of  13'  in  the  angle  we 
assume  that  there  will  be  an  increase  of  Jjj-  of  .014  which  is 
.003.  Adding  this  to  .574,  we  have  .577.  Therefore 
sin  35°  13'  =  .577. 

Example  2.     To  find  the  cos  72°  44'. 

Since  this  angle  is  more  than  45°,  run  up  the  last  column 
until  you  come  to  72°.  Look  along  the  bottom  line  until  you 
come  to  the  column  marked  cosine.  In  this  column  in  a  line 
with  72°,  you  will  find 

cos  72°  =  .309. 

Also  cos  73°  =.292. 

Our  angle  lies  between  72°  and  73°.     Our  argument  now  is, 
since  as  the  angle  increases  60',  its  cosine  decreases  .017, 


VI,  §  238]  TRIGONOMETRIC  FUNCTIONS  207 

therefore  as  the  angle  increases  44  '  the  cosine  will  decrease  -j$ 
of  .017,  which  is  .012.  Subtracting  this  from  .309,  we  have 
.297. 

Therefore  cos  72°  44'  =  .297. 

When  the  remainder  in  taking  the  fractional  part  for  the 
increase  or  decrease  is  more  than  .0005,  we  use  it  as  .001. 
When  it  is  less  than  .0005  we  do  not  use  it.  Do  not 
carry  out  the  decimal  farther  than  the  table  that  you  are 
using. 

Exercise.     Find  the  following: 

sin  54°  28'.  cos  63°  32'. 

cos  22°    8'.  tan  56°  58'. 

tan  47°  34'.  sin  15°  18'. 

238.  The  next  work  is  to  see  how  from  the  tables  we  can 
find  the  number  of  degrees  and  minutes  in  an  angle  if  we 
know  its  function. 

Example  1.     If  sin  <f>  =  .475,  find  <f>. 

Running  down  the  column  marked  sin  0,  we  find 
sin  28°  =  .469. 
sin  29°  =  .485. 

Since  .475  lies  between  these  functions,  the  angle  must  be  be- 
tween these  angles.  So  our  argument  is,  since  as  'the  func- 
tion increases  .016,  our  angle  increases  60',  therefore  as  the 


function  increases  .006,  the  angle  will  increase  '^y^  of  60', 

which  is  |  of  60',  which  is  22'.     Therefore 

0  =  28°  22'. 

Example  2.     If  cos  <£  =  .621,  find  <£. 
Looking  in  the  column  marked  cos  <f>,  we  find 
cos  51°  =  .629. 
cos  52°  =  .616. 

Since  .621  lies  between  the  two  functions,  our  angle  must 
lie  between  these  two  angles.  So  our  argument  is:  When  the 
function  decreases  .013,  our  angle  increases  60',  therefore 


208 


PLANE  GEOMETRY 


[VI,  §  239 


f\f\Q 

when  our  function  decreases  .008,  our  angle  increases  '-^—^  of 

60'  which  is  }  of  60',  which  is  30'.  Therefore  0  =  51°  30'. 
Find  <£  in  the  following: 

sin  4>  =  0.885.  cos  <j>  =  0.40. 

cos<£  =0.969.  tan0  =  0.79. 

tan<£  =  8.00.  sin  <j>  =  0.521. 

239.  Now  that  we  are  able  to  find  these  six  ratios  from  our 
table,  we  have  six  equations  which  we  can  use  as  formulas, 
when  we  have  taken  proper  measurements. 

The  following  simple  device  is  suggested  for  measuring 
angles,  when  a  surveyors'  transit  cannot  be 
had. 

Fasten  together  two  yard  sticks  or  foot 
rules  so  that  they  will  turn  as  on  a  hinge. 
To  the  rivet  hang  a  string  with  a  weight  on       / 
it  (a  plumb  line).     With  thumb-tacks  fasten  </ 
your  triangular  protractor  to   one  of   the 
sticks,  the  long  side  running  along  the  stick, 
and  its  center  directly  on  the  center  of  the 
rivet. 

To  determine  the  height  of  an  object,  as  AiB  in  the  figure 
below,  hold  the  instrument  at  0  and  measure  Z  A  OB.  A 
stick  or  pole  should  be  used  to  support  the  instrument,  and 
the  distance  OiO  should  be  measured,  to  be  added  to  AB. 

The  distance  OiAi  which  is  equal  to  OA  can  be  measured. 
Thus  we  shall  have  <j>  and  x,  to  8 

find  y  in  the  formula, 

—  =  tan  <f>,         y  =  x  tan  0. 

Example  1.     Suppose  that  an- 
gle <f>  is  23°  16',  and  distance  OA 


is  300  feet. 
of 


What  is  the  height   ° 


vi,  §  239]  TRIGONOMETRIC  FUNCTIONS  209 

Given  x  =  300,  and  0  =  23°  16'.     To  find  y. 
Selecting  the  function  which  contains  the  parts  we  have 
given  and  also  the  part  which  we  are  trying  to  find,  we  choose 

y 

tan  0  =  -. 
x 

Solving  for  y,  y  =  x  tan  4>. 

Substituting  values:  y  =  300.tan23°  16'. 

y  =  300  X  0.430  (finding  value  of  tan  23° 

16  'from  the  table). 
=  129.0  number  of  feet  in  height  AB. 

Check  this  by  making  a  careful  drawing  on  cross  section 
paper  and  comparing  results. 

Example  2.  Suppose  that  you  wished  to  find  the  distance 
from  one  point  to  another  between  which  there  is  an  obstruc- 
tion. The  following  measurements  are  suggested. 

Let  the  points  be  P  and  Q.  Adjust  your  sticks  at  right 
angles  with  the  center  of  the  protractor  on  point  P,  and  one 
of  the  arms  pointing  in  the  direc- 
tion  of  Q.  From  P  in  the  direction 
of  the  other  arm  measure  any  con- 
venient length  PO.  At  point  0  ad- 
just your  sticks  so  that  one  will  point 
in  the  direction  of  point  P  and  the  c 
other  in  the  direction  of  point  Q.  Read  the  measurement  of 
the  angle  at  0. 

Suppose  that  the  distance  OP  is  150  feet  and  the  angle  at 
0  is  52°  29'.  What  is  the  distance  PQ1 

Let  x  =  the  measured  distance  OP. 

Let  y  =  the  required  distance  PQ. 

Given  x  =  150,  and  0  =  52°  29'.     To  find  y. 

Selecting  the  function  which  contains  the  given  parts  and 
the  parts  that  we  wish  to  find: 


Solving  for  y,  y  =  x  tan  <f>. 


210  PLANE  GEOMETRY  [vi,  §  240 

Substituting  the  given  values,  y  =  150  X  tan  52°  29 '. 
Finding  value  of  tan  0,  y  =  150  X  1.302. 

=  195.3   number  of  feet  in 

distance  PQ. 

Check  by  making  careful  drawing  on  cross-section  paper 
and  comparing  results. 

Example  3.  Suppose  that  you  cannot  go  to  the  object  whose 
height  you  wish  to  know.  J> 

The  following  figure  sug- 
gests measurements  to  take: 
AB  =  20  chains. 
0  =  25°. 

&  =40°.  -*>  --* m" 

Graphic  Solution.  On  your  paper  lay  off  a  distance  20 
units.  At  each  end  draw  the  measured  angles  as  shown  in  the 
figure.  The  arms  will  meet  at  point  D.  Prolong  A B  and 
drop  a  perpendicular  from  D.  Measure  CD. 

Solution  by  Functions.  Let  the  unknown  distance  EC  be 
m  chains  and  let  CD  be  y  chains.  Then  we  have: 

From  A  BCD,  ^  =  tan  40°. 

From  A  ACD,  2Q  +  m  =  tan  25°* 

Solve  for  y  and  m. 

20  tan  25° 

Ans.  m  = 


tan  40°  -tan  25° ' 
20  tan  25°  tan  40° 
y  "tan  40°  -tan  25°' 

The  above  are  given  as  suggestions.  The  student  can 
work  out  many  ways  of  finding  distances  which  cannot  be 
measured  directly. 

240.  Exercises.  In  the  following  make  drawings,  solve 
for  the  parts  called  for  by  making  use  of  the  trigonometric 
functions  and  check  by  the  drawings. 


VI,  §240]  TRIGONOMETRIC  FUNCTIONS  211 

Here  x,  y,  r,  (f>  are  used  as  in  the  figure  of  Art.  234. 

1.  x  =  27.3,  0  =  59°  13 ',  find  y  and  r. 

2.  r  =  1.57,  y  =  1,  find  0  and  x. 

3.  ?/  =  27.33,  r  =  67.1,      find  <£  and  x. 

4.  ?/  =  256,  0  =  35°  57',  find  a;  and  r. 

The  following  problems  are  given  to  show  the  uses  of 
trigonometric  functions.  The  best  problems  for  the  student  to 
solve  are  those  for  which  he  has  done  his  own  measuring,  as 
they  have  a  personal  interest. 

5.  At  a  horizontal  distance  of  120  feet  from  the  foot  of  a 
steeple  the  angle  of  elevation  of  the  top  of  the  steeple  is 
60°  33'.     Find  the  height  of  the  steeple. 

6.  A  train  is  running  at  the  rate  of  30  miles  an  hour.     The 
mail  pouch  is  thrown  from  it  at  right  angles  to  the  direction 
of  the  train.     The  pouch  takes  a  direction  which  makes  an 
angle  of  30  degrees  with  the  direction  of  the  train.     With 
what  speed  was  the  pouch  thrown? 

7.  Two  forces  act  at  right  angles  to  one  another.     One  is 
20  Ibs.  and  the  other  is  35  Ibs.     What  angle  will  the  resultant 
force  make  with  the  greater  of  the  two  forces? 

8.  Two  forces  acting  at  right  angles  give  a  resultant  force 
of  32.5  pounds.     If  one  of  the  forces  makes  an  angle  of 
27°  38'  with  the  resultant  force,  what  is  the  number  of  pounds 
in  each  force? 

9.  Two  forces,  of  which  the  first  is  47  pounds  less  than  the 
second,  act  at  right  angles  to  each  other.     Their  resultant 
force  is  65  pounds.     Find  the  number  of  pounds  in  each 
force,  and  the  angle  which  the  resultant  force  makes  with  each, 

10.  Rain-drops  are  falling  vertically  with  a  speed  of  100 
miles  an  hour  as  they  pass  the  windows  of  a  train  which  is 
running  30  miles  an  hour.     How  will  they  seem  to  fall  to  a 
person  in  the  train? 

11.  The  tower  of  Pisa  is  172  feet  high  and  inclines  11  feet 
and  2  inches  from  the  perpendicular.     What  is  the  angle  of 
inclination? 


212  PLANE  GEOMETRY  [VI,  §  241 

241.  Ratio  and  Proportion  Involving  Variables. 

Note.  The  student  may  review  First  Course,  pages  245 
to  255,  before  going  on  with  the  work. 

One  of  the  simplest  statements  of  the  relation  of  two 
variables  is 

or,  in  ratio  form,  —  =  a. 

If  we  regard  x  and  y  as  the  coordinates  of  a  variable  point, 
this  statement  calls  for  all  points  such  that  the  ratio  of 
ordinate  to  abscissa  is  the  fixed  number  a. 

In  mathematical  language  we  are  to  find  the  locus  of  points 
whose  coordinates  satisfy  the  equation  P"  ,N 

^  =  a. 

x 

To  prove  that  this  locus  is  a  straight  line. 

Locate  point  P  by  counting  x  =  1,  and 
y  =  a.  This  is  one  point  of  the  locus. 
Draw  a  straight  line  MN  through  this 
point  and  the  origin,  which  is  another 
point  of  the  locus. 

To  prove  that  MN  is  the  locus  of  points  whose  coordinates 
will  always  have  the  ratio  a. 

Analysis.  In  order  to  prove  that  this  line  is  the  locus 
required,  we  must  prove  that  any  point  on  the  line  has  co- 
ordinates which  satisfy  the  equation,  and  that  any  point  not 
on  the  line  has  coordinates  which  will  not  satisfy  the  equation. 

Let  Pr  be  any  other  point  on  the  line  other  than  point  P. 
Let  its  coordinates  be  x',  y'. 

y' 

We  must  prove  that  —f  =  a. 

Proof.  The  triangles  P'OR  and  OQP  are  similar.  Prove 
this. 

Therefore  y' :  x'  =  a  :  1.  Why? 


Q   T 


VI,  §242]  TRIGONOMETRIC  FUNCTIONS  213 

Therefore  the  coordinates  of  P'  satisfy  the  requirement 
that  they  have  the  ratio  a. 

Let  P"  be  any  point  not  on  the  line  MN.  Let  its  coordi- 
nates be  x",  y". 

y" 
We  must  prove  that  — 7-,  does  not  equal  a. 

3C 

Let  y"  cut  line  MN  at  point  S.  The  coordinates  of  point 
SareO",  ST). 

am 

Since  point  S  is  on  the  line  MN:          —r,  =  a.  Why? 

JU 

Therefore  ^77  does  not  equal  a  umess  y"  equals  ST.  (Why?) 

JU 

Therefore  the  coordinates  of  P"  do  not  satisfy  the  condi- 
tion that  their  ratio  is  a. 

So  we  have  proved  the  theorem: 

242.  Theorem  XXII.     The  graph  of  the  equation  y  =  ax  is 
a  straight  line  through  the  origin  and  point  (1,  a). 

243.  Theorem  XXIII.     The  graph  of  the  equation  y  =  ax 
+  b  is  a  straight  line. 

This  equation  is  but  the  statement  that  for  every  value  of 
x  the  value  of  y  is  6  units  more  than  the  value  of  y  in  the  equa- 
tion y  =  ax. 

Since  the  graph  of  y  =  ax  is  a  straight  line,  the  graph  of 
y  =  ax  +  b  is  a  straight  line  parallel  to  it,  and  b  units  higher 
or  lower,  depending  on  the  sign  of  6. 

Exercises.     Draw  the  graph  of  the  following: 
1.  y  =  Jx.  2.  y  =Jz  +  3. 

3.  y  =  3  x.  4.  y  =  3  x  -  5. 

6.  y  —  \x  —  1.         6.  2x  —  2y  =  5  (rewrite  as  y  =  x—  2J). 

7.  -3x  +  7y  =  14.  8.  4x-2y  =  -7. 

9.  A  tree,  now  6  inches  in  diameter,  grows  at  the  rate  of  a 
inches  a  year.  How  thick  will  it  be  in  x  years?  If  the  thick- 
ness is  y,  show  that  y  =  ax  +  b.  Draw  the  graph  when  a  = 
0.5  and  6=4. 


214 


PLANE  GEOMETRY 


[VI,  §  244 


244.  Since  the  inclination  of  the  line  y  =  ax  +  b  to  the  axis 
of  the  abscissas  is  the  same  as  that  of  the  line  y  =  ax,  we  can 
readily  find  the  angle  which  the  graph  of  any  linear  equation 
makes  with  the  axis  of  abscissas.  Since  for  any  straight  line 
through  the  origin  we  have 


and  since  tan  6  =— , 

x 

therefore  tan  <£  =  a. 

So  that,  knowing  a,  we  can  find  value  of  </>  from  the  tables. 

Definition.  The  value  of  tan  <j> 
is  called  the  slope  of  the  line. 
Example.    Take  the  line  y  = 


This  is  parallel  to  the  line  y  = 
\x. 

Therefore      tan  0  =  J.      Why? 
0  -  26°  34'. 

Therefore  the  graph  of  y  = 
J  x  —  3  makes  an  angle  of  26°  34 '  with  the  axis  of  the  abscissa. 

Exercises.     Determine  the  angle  which  the  graphs  of  the 
following  equations  make  with  the  axis  of  abscissas. 

1.  2  x  -  3  y  =  7.  Rewrite  as  y  =  f  x  -  J. 

2.  -  5  x  +  y  =  1.  4.  7  x  -  y  =  -  14. 

3.  4  x  -  16  y  =  48. 

245.  Angle  between  Two  Lines. 
If  two  lines  are  drawn  with  reference 
to  the  same  axes,  we  can  easily  find 
the  angle  which  they  make  with  one 
another.  In  the  adjacent  figure  the 
lines  are  the  graphs  of  the  equations 

x-2y  =  -4, 

2x-y  =  1. 


C^ 


y. 


ASL 


VI,  §  245]  TRIGONOMETRIC  FUNCTIONS  215 

Let  <f>  be  the  angle  which  graph  of  x  —  2  y  =  —  4  makes 
with  the  axis  of  abscissas,  and  let  fa  be  the  angle  which  the 
graph  of  2  x  —  y  =  I  makes  with  the  axis  of  abscissas. 

Then  fa  —  <£  is  the  angle  between  the  two  lines.     (Art.  65.) 

Find  the  angle  which  the  lines  of  the  above  exercises  make 
with  one  another,  by  first  finding  the  angles  </>  and  fa.  Check 
by  measurement  of  an  accurately  drawn  figure. 

When  tan  </>  is  positive,  that  is,  when  a  is  positive  <f>  is  acute 
and  can  easily  be  found  from  the  tables. 

When  tan  <£  is  negative,  we  shall  have  to  examine  a  little 
farther,  since  there  are  no  negative  numbers  in  our  table. 


We  have  given  the  line  MN  making  the  angle  $  with  the 
axis  of  the  abscissa,  <£  being  greater  than  a  right  angle.  We 
can  find  an  angle  less  than  a  right  angle  which  has  a  tangent 
whose  absolute  value  is  the  same  as  that  of  angle  <f>. 

Analysis.  In  order  to  do  this  we  draw  through  the  origin 
a  line  parallel  to  MN.  On  it  mark  point  P.  Draw  angle  fa 
equal  to  the  supplement  of  <j>.  Make  OPi  equal  to  OP. 

From  this  it  is  readily  proved  that 

y  =  2/1,    and     x  =  -  XL 

Therefore  £«_!£.  Why? 

X  Xi 

That  is  tan  <j>  =  —  tan  fa. 

We  have  shown  that  the  tangent  of  an  angle  is  equal  to 
the  negative  tangent  of  its  supplement.  To  find  an  angle 
whose  tangent  is  negative,  take  the  supplement  of  the  angle 
whose  tangent  is  numerically  the  same,  but  positive. 


216 


PLANE  GEOMETRY 


[VI,  §  246 


246.  Your  attention  is  especially  called  to  the  use  of  the 
negative  sign  in  the  above.     Students  are  too  much  inclined 
to  say  that  an  expression  is  a  negative  number  because  they  see 
a  negative  sign  written  before  it.     In  the  expression  x  =  —  Xi, 
by  the  figure  you  see  readily  that  Xi  is  not  negative.     Neither 
does  this  algebraic  expression  say  that  it  is  negative.    In  fact  it 
states  nothing  as  to  the  real  direction  of  either  x  or  x\.     It 
merely  states  that  x  is  of  the  opposite  direction  to  x\.     The 
student  should  fix  in  mind  this  fact,  that  when  a  negative  sign 
precedes  a  general  number,  it  does  not  indicate  that  the  num- 
ber is  negative,  but  that  the  expression  is  the  negative  of  what- 
ever particular  value  may  be  given  to  the  general  number. 

tan  4>  =  —  tan  fa  does  not  state  that  tan  fa  is  negative. 
We  see  that  in  the  above  figure,  tan  fa  is  positive.  The  ex- 
pression states  that  tan  <£  is  the  negative  of  tan  fa.  Compare 
the  other  functions  of  angle  </>  with  those  of  fa. 

247.  Negative  Slope.     Given  the  equation 

4  x  +  3  y  =  15. 

To  make  the  graph  and  find  its  inclination  to  the  axis  of 
the  abscissa. 

Writing  this  equation  in  the  form  y  =  ax  +  b,  we  have 

4 


The  equation  of  the  line  through  the  origin  parallel  to  the 
graph  of  this  line  is 

4^/4 


Draw  the  graph  of  this  by 
counting  —  3  in  the  direction  of 
the  axis  of  the  abscissa,  and  4  in 
the  direction  of  the  axis  of  the 
ordinate.  Draw  a  line  through 
this  point  and  the  origin. 

Since  for  every  point  on  the 


VI,  §  248]  TRIGONOMETRIC  FUNCTIONS  217 

4 

graph  of  the  equation  y  =  —  -  #  +  5,  yis  5  units  more  than 

o 

for  the  corresponding  point  on  the  line  parallel  to  it  through 
the  origin,  for  the  point  corresponding  to  the  origin  we  locate 
the  point  (0,  5),  and  for  the  point  (—  3,  4),  we  locate  (—  3,  9). 
Draw  a  line  through  these  points  and  we  have  the  graph  of  the 
given  equation. 

To  find  the  value  of  angle  <f>. 
From  the  equation  we  have 

y__       £ 
x~    "  3' 

11  4 

that  is  tan  <£  =  —  =  —  -5- 

x  o 

=  -  1.33+. 

Looking  in  the  table  for  the  angle  whose  tangent  is  1.33, 
we  find  tan  53°  =  1.33.  Since  the  tangent  of  the  angle  is  the 
negative  of  the  tangent  of  its  supplement,  tan  127°  =  —  1.33. 

Therefore  the  line  makes  an  angle  of  127°  with  the  axis  of 
the  abscissa. 

248.  Exercises.  In  the  following  make  graphs,  find  the 
angle  which  each  line  makes  with  the  axis  of  the  abscissa, 
and  the  angle  which  they  make  with  one  another.  Solve  for 
x  and  y  by  algebra  and  note  correspondence  of  values  of  x 
and  y  with  coordinates  of  point  of  intersection  of  the  lines. 

1.  2x  +  3y  =7,  2.  2x-3y  =  10, 

33  +  40  =  10.  5x  +  2y  =6. 

3.  2  x  -  10  y  =  15,  4.  3  x  +  y  =  4, 

2z-4y  =  18.  x  +  3y  =  -2. 

6.  3  -  15  y  =  -  xt  6.  6  y  -  10  x  =  14, 
3  -  15y  =4  s.  y  -x  =3. 

7.  2 y  +  5=7 x,  8.    -5  -3x  =5y 
x  +  3     =  -42.  2x  +     =6. 


218  PLANE  GEOMETRY  [VI.  §  249 

249.  Summary  and  Questions. 
Part  I — Ratio  and  Proportion. 

Definition.  Ratio,  proportion,  commensurable  and  incommensurable 
numbers,  antecedent,  consequent,  mean  proportional. 

Pencil  of  lines. 

Experiment  by  measurement  to  bring  out  theorems. 

Theorem  I.  If  a  pencil  of  lines  is  cut  by  a  pencil  of  parallels,  the 
corresponding  segments  are  in  proportion. 

Remark.  On  the  generalization  of  this  theorem  for  incommensurable 
cases. 

Theorem  II.  If  four  quantities  are  in  proportion,  they  are  in  pro- 
portion by  inversion. 

Theorem  III.  If  four  quantities  are  in  proportion,  they  are  in  propor- 
tion by  alternation. 

Theorem  IV.  In  a  series  of  equal  ratios  the  sum  of  an  antecedent  and 
its  consequent  is  to  the  consequent  as  the  sum  of  any  other  antecedent 
and  consequent  is  to  the  consequent. 

Theorem  V.  In  a  series  of  equal  ratios  the  difference  between  any 
antecedent  and  its  consequent  is  to  the  consequent,  as  the  difference 
between  any  other  antecedent  and  consequent  is  to  the  consequent. 

Theorem  VI.  If  four  quantities  are  in  proportion  the  sum  of  an  ante- 
cedent and  its  consequent  is  to  their  difference  as  the  sum  of  the 
other  antecedent  and  consequent  is  to  their  difference. 

Theorem  VII.  In  a  series  of  equal  ratios  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

Problems  I  and  II.     To  divide  a  line  in  any  given  ratio. 

Definition  of  internal  and  external  division. 

Theorem  VIII.  If  a  pencil  of  lines  is  cut  by  a  pencil  of  parallels 
the  segments  of  the  lines  of  the  pencil  (cut  from  the  vertex)  are  pro- 
portional to  the  corresponding  segments  of  the  parallels. 

Theorem  IX.  If  a  pencil  of  lines  is  cut  by  a  pencil  of  parallels,  the 
corresponding  segments  of  the  pencil  of  parallels  are  in  proportion. 

Theorem  X.  A  line  parallel  to  the  side  of  a  triangle  divides  the  other 
two  sides  proportionally. 

Theorem  XI.  A  line  which  divides  two  sides  of  a  triangle  propor- 
tionally is  parallel  to  the  third  side. 

Theorem  XII.  If  an  angle  of  a  triangle  is  bisected  by  a  line  cutting 
the  opposite  side,  then  the  opposite  side  is  divided  internally  into  seg- 
ments proportional  to  the  other  two  sides. 


VI.  §  249]  TRIGONOMETRIC  FUNCTIONS  219 

Theorem  XIII.  If  the  exterior  angle  of  a  triangle  is  bisected  by  a 
line  cutting  the  opposite  side,  then  the  opposite  side  is  divided  externally 
into  segments  proportional  to  the  other  two  sides. 

Part  II — Similar  Figures. 

Definition.     Similar  systems  of  points  and  similar  figures. 

Corollary  1.  If  two  triangles  are  similar  their  corresponding  sides  are 
proportional. 

Corollary  2.  If  two  triangles  are  similar  they  are  mutually  equi- 
angular. 

Theorem  XIV.  Two  triangles  are  similar  if  they  have  two  angles 
of  the  one  equal  respectively  to  the  two  angles  of  the  other. 

Corollary  1.  Two  triangles  are  similar  if  their  corresponding  sides 
are  parallel  each  to  each,  or  are  perpendicular  each  to  each. 

Corollary  2.     Equilateral  triangles  are1  similar. 

Corollary  3.  Two  right  triangles  are  similar  if  they  have  an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other. 

Theorem  XV.  Corresponding  altitudes  of  similar  triangles  are  pro- 
portional to  any  two  corresponding  sides. 

Theorem  XVI.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle 
of  the  other,  and  the  including  sides  proportional,  the  triangles  are  similar. 

Theorem  XVII.  If  two  triangles  have  their  corresponding  sides  pro- 
portional they  are  similar. 

Theorem  XVIII.  If  two  polygons  are  similar  they  can  be  divided 
into  the  same  number  of  similar  triangles,  similarly  placed. 

Corollary  1.  If  two  polygons  are  similar,  they  have  their  corre- 
sponding sides  proportional  and  their  corresponding  angles  equal. 

Corollary  2.  The  perimeters  of  similar  polygons  have  the  same  ratio 
as  their  corresponding  sides. 

Theorem  XIX.  If  two  polygons  are  mutually  equiangular  and  have 
their  corresponding  sides  proportional,  they  are  similar. 

Theorem  XX.  If  a  pencil  of  lines  is  cut  by  a  circumference,  the 
product  of  the  segments  (cut  from  the  vertex)  is  constant. 

Corollary.  A  tangent  is  a  mean  proportional  between  the  segments 
cut  on  any  other  ray  of  the  pencil. 

Theorem  XXI.  If  a  perpendicular  is  dropped  from  the  vertex  of  the 
right  angle  of  a  right-angled  triangle,  to  the  hypotenuse, 

(a)  it  divides  the  triangle  into  two  similar  triangles  each  similar 
to  the  original  triangle; 

(b)  either  side  of  the  triangle  is  a  mean  proportional  between  the 
hypotenuse  and  the  adjacent  segment  into  which  the  hypotenuse  is 
divided; 


220  PLANE  GEOMETRY  [VI,  §  249 

(c)  the  perpendicular  is  a  mean  proportional  between  the  segments 
into  which  the  hypotenuse  is  divided. 

Problem  V.     To  construct  a  mean  proportional  between  two  given 
line  segments. 

Corollary.     To  construct  a  square  equal  to  a  given  rectangle. 

Part  III — Trigonometric  functions. 

Opening  discussion  and  illustration. 

Definition  of  trigonometric  functions. 

Tables.     Explanation  of  use,  and  exercises  in  use. 

Ratio  and  Proportion  involving  variables. 

Theorem  XXII.     The  graph  of  the  equation  y  =  ax  is  a  straight  line 
through  the  origin  and  point  (1,  a). 

Theorem  XXIII.     The  graph  of  a  linear  equation  is  a  straight  line. 

1.  Into  what  three  parts  is  this  chapter  divided? 

2.  What  is  the  ratio  of  one  number  to  another. 

3.  Illustrate  commensurable  and  incommensurable  ratios  of  line- 
segments. 

4.  State  some  theorems  which  may  arise  when  you  have  four  quan- 
tities in  proportion. 

5.  Explain  how  they  are  proved  by  proving  that  if  a  pencil  of  lines 
is  cut  by  a  pencil  of  parallels,  the  corresponding  segments  are  in  pro- 
portion. 

6.  Describe  what  is  meant  by  similar  systems  of  points. 

7.  What  are  similar  triangles? 

8.  State  the  theorems  on  congruent  triangles  and  similar  triangles  in 
corresponding  pairs. 

9.  Are  there  any  propositions  in  this  chapter  which  were  proved 
without  using  any  of  the  preceding  theorems?     If  so,  state  them. 

10.  Give  the  six  trigonometric  functions  and  tell  why  they  are  called 
the  functions  of  an  angle. 

11.  To  what  practical  use  can  the  trigonometric  functions  be  put? 

12.  Do  you  regard  the  work  of  this  chapter  any  more  practical  than 
that  of  the  preceding  chapter? 

13.  Look  through  the  list  of  exercises  and  select  those  that  you 
regard  as  practical  in  nature. 

14.  What  is  the  graph  of  a  linear  equation? 

Most  of  the  theorems  of  this  chapter  are  selected  from  Book  IV  of 
Euclid's  Geometry, 


CHAPTER  VII 


PART  I— AREAS  OF  RECTILINEAR  FIGURES.  MEN- 
SURATION OF  CIRCLES.  PART  II— DIVISION  OF 
A  PERIGON.  PART  III— INCOMMENSURABLE 
CASES 

PART  I— AREAS  OF  RECTILINEAR  FIGURES 

250.  Theorem  I.     Two  rectangles  having  equal  altitudes  are 
to  each  other  as  their  bases. 


•R: 


Given  the  rectangles  R  and  Rf  with  equal  altitudes  a, 
and  with  unequal  bases  6  and  &',  respectively. 

.  area  of  R        b 
To  prove  that  -         Jf  p/  =  ^ 
area  of  R'       b 

Analysis.  We  can  prove  this  theorem  by  expressing  the 
areas  of  R  and  R'  in  terms  of  an  area-measuring  unit,  and 
the  lengths  of  b  and  &'  in  terms  of  a  linear  measuring  unit, 
and  showing  their  ratios  to  be  equal.  To  do  this  divide  6 
and  b'  by  the  same  unit  of  measure  u.  Let  it  go  m  times  into 
b,  and  n  times  into  b'.  At  the  points  of  division  draw  lines 
perpendicular  to  6;  these  will  divide  R  into  m  equal  rectangles, 
and  R'  into  n  equal  rectangles,  whose  area  is  r.  Now  by 
finding  the  lengths  of  b  and  b'  in  terms  of  u,  and  from  this 
finding  their  ratio,  then  by  finding  the  areas  of  R  and  R'  in 
terms  of  r}  and  finding  their  ratio,  we  can  prove  the  propor- 
tion in  the  theorem. 

221 


222  PLANE  GEOMETRY  [vn,  §  250 

Proof.  The  rectangles  iflto  which  R  and  Rf  are  divided 
are  equal.  Why? 

b  =  mu, 

and  6'  =  nu.  Why? 

Area  of  R  =  rar, 
Area  of  R'  =  nr.  Why? 

0     ,                                      6      mw      m  TT7,    0 

So  that  ^7  = = —  Why? 

o        nu      n 

Area  of  R        mr      m  „,,    0 

Also,  Jf  p,  =  —  =  —  Why? 

Area  of  R'       nr       n 

Area  of  R        b_ 
Area  of  R'      b' 

State  the  theorem  proved. 

Discussion.  In  the  figure,  and  in  the  proof,  it  was  assumed 
that  b  and  b'  have  a  common  unit  of  measure. 

This  proof  would  not  hold  if  the  base  of  one  rectangle  were 
equal  to  the  side  of  a  square  and  the  base  of  the  other  rect- 
angle were  equal  to  the  diagonal  of  the  same  square.  Why 
not? 

Neither  could  we  use  this  proof  if  the  base  of  one  rectangle 
were  equal  to  the  circumference  of  a  circle  and  the  base  of 
the  other  were  equal  to  the  radius  of  the  same  circle.  Why 
not? 

The  proof  that  this  theorem  is  true  for  such  cases,  com- 
monly known  as  the  incommensurable  cases,  is  given  after 
the  treatment  of  theory  of  limits  at  the  end  of  this  chapter. 
When  this  theorem  is  proved  for  the  incommensurable  cases, 
all  the  proofs  of  following  theorems  which  depend  upon  this 
theorem  will  hold  for  incommensurable  cases  without  further 
discussion. 

Corollary  1.  Rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Corollary  2.  Parallelograms  having  equal  altitudes  are  to 
each  other  as  their  bases,  and  having  equal  bases  are  to  each 
other  as  their  altitudes.  Art.  94,  Cor.  2. 


VII,  §  251] 


MENSURATION 


223 


Corollary  3.  Triangles  having  equal  altitudes  are  to  each 
other  as  their  bases,  and  having  equal  bases  are  to  each  other  as 
their  altitudes. 

251.  Theorem  II.  Two  rectangles  are  to  each  other  as  the 
product  of  their  bases  and  altitudes. 


ti 

ft' 

R" 

h' 

R 

Given  two  rectangles  R  and  R'y  with  bases  b  and  bf  and 
altitudes  h  and  h'. 

area  of  R         bh 


To  prove  that 


area  of  R'       Vh' 


Analysis.  Since  the  only  propositions  that  we  have  proved 
deal  either  with  rectangles  of  equal  bases  or  of  equal  altitudes, 
we  shall  form  such  rectangles.  We  can  get  these  by  drawing 
a  new  rectangle  R",  whose  base  is  6  and  whose  altitude  is  h'. 
Now  by  writing  the  ratios  of  each  of  the  given  rectangles  to 
this  new  rectangle,  we  can,  by  dividing,  eliminate  the  new 
rectangle  and  have  the  ratio  called  for  in  the  proposition. 

4=  p.  Why? 


Pr00f' 


_ 
R"b 


Dividing  the  first  equation  by  the  second, 


Why? 


Why? 


R'~Vh'' 
State  the  theorem  proved. 

Exercise.     Find  the  ratio  of  the  rectangles  whose  dimen- 
sions are: 

(a)  2  by  3,  and  5  by  6.  (6)  5  by  \/2,  and  2  V2  by  20. 


224  PLANE  GEOMETRY  [vn,  §  252 

252.  Theorem  III.  The  area  of  a  rectangle  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Note.  By  this  we  mean  that  the  number  of  square  units  in  the  area 
is  equal  to  the  number  of  linear  units  in  the  base  multiplied  by  the 
number  of  linear  units  in  the  altitude. 


Given  the  rectangle  R,  with  base  b  units  long  and  altitude 
h  units. 

To  prove  that  the  area  of  R  equals  bh. 

Analysis.  To  find  the  area  of  R  we  must  find  its  ratio  to 
a  unit  of  area;  we  select  the  square  unit  U  whose  dimension 
is  1.  We  have  now  but  to  find  the  ratio  of  rectangle  R  to 
rectangle  U. 

Proof.  jj  =  — .    According  to  Theorem  II. 

The  ratio  of  R  to  U  is  the  area  of  R; 
area  of  R  =  bh. 

State  the  theorem  proved. 

Corollary  1.  The  area  of  a  parallelogram  is  equal  to  the  prod- 
uct of  its  base  by  its  altitude. 

Corollary  2.  The  area  of  a  triangle  is  equal  to  one-half  the 
product  of  its  base  by  its  altitude. 

Corollary  3.  The  area  of  a  trapezoid  is  equal  to  one-half  the 
product  of  the  sum  of  its  bases  by  its  altitude. 

253.  Exercises.  (Draw  careful  figures  on  cross-ruled 
paper.) 

1.  Find  the  area  of  the  triangle  whose  vertices  are  the 
points  (-  3,  2),  (1,  2),  (},  5). 

2.  What  kind  of  a  figure  is  formed  by  the  points  (—  1,  —  3), 
(7,  -  3),  (9,  4),  (1,  4).    Compute  the  area. 


VII,  §  253] 


MENSURATION 


225 


3.  What  kind  of  a  figure  has  for  its  vertices  the  points 
(-  10,  --  7),   (12,  --  7),   (5,  2),  (-  2,  2)?     Compute  the 
area. 

4.  Compute  the  area  of  the  figure  the  coordinates  of  whose 
vertices  are  (-  10,  -  7),  (12,  -  7),  (5,  2),  (1,  3),  (-  2,  2). 

5.  Compute  the  area  of  the  figure  whose  vertices  are  the 
points  (-  10,  -  7),  (14,  9),  (12,  -  7),  (5,  2),  (1,  3),  (-  2,  2). 

6.  How  many  acres  in  this  field  if  the  side  of  one  small 
square  is  10  rods? 


Do  this  by  starting  with  the  rectangle  whose  corners  are 
marked  by  the  dots  and  cutting  off  triangles  until  you  have 
left  the  area  required. 

Study  out  other  methods  for  getting  this  area. 

7.  What  is  the  altitude  of  an  equilateral  triangle  whose 
perimeter  is  15  units  and  area  60  square  units?     Solve  when 
the  perimeter  is  p  units  and  the  area  a  square  units. 

8.  In  an  isosceles  trapezoid  the  altitude  is  12  units  and 
one  of  the  equal  sides  is  13  units.     What  is  its  area?     Solve 
when  the  altitude  is  a  units  and  one  of  the  equal  sides 
is  s  units. 

9.  A  rectangle  which  is  twice  as  wide  as  it  is  high  is  in- 
scribed in  a  circle  of  radius  10  units.     Calculate  the  dimen- 
sions and  the  area  of  the  rectangle.     Solve  when  the  base  is 
n  times  the  height,  and  the  radius  is  r. 


226  PLANE  GEOMETRY  [vn,  §  254 

254.  Theorem  IV.  Two  similar  triangles  are  to  each  other 
as  the  squares  of  two  corresponding  sides,  or  as  the  squares  of 
two  corresponding  altitudes. 


Given  triangles  ABC  and  A'B'C'^  with  the  correspond- 
ing sides  a  and  a',  6  and  &',  c  and  c',  and  altitudes  h  and  h'. 

area  of  A  ABC         a*        b2        c2        h2 
To  prove  that  ma  of  A  A^c  =  ft  =  y,  =  ^  =  FI- 


Analysis.  Since  the  propositions  which  we  have  had  on 
areas  of  triangles  dealt  with  the  bases  and  altitudes,  we  use 
these.  Now  by  writing  the  area  of  each  triangle  and  finding 

the  ratio  between  them,  and  remembering  that  7-7  =7-7  =—  =  — , 

it      u      a      o 

we  shall  be  able  to  prove  the  equality  of  the  ratios  called  for 
in  the  theorem.  Let  T  denote  the  area  of  A  ABC,  and  T' 
the  area  of  A  A'B'C'. 

Proof.  T  =  \  bh.  Why? 

T  =  i  b'h'.  Why? 

T       \bh        bh  _    9 

T  =  Wh'  =  vh"  my? 

But  since  the  triangles  are  similar, 

£_£_£_£.    (Art.  221.) 

Substituting  in  the  above  equation,  we  have 

_±_  — _._—_  —  _._fL._.5L  wviir? 

T'~b'     hr      b'2     h'2     a/2~  cf2' 

State  the  theorem  proved. 


VII,  §  255] 


MENSURATION 


227 


255.  Theorem  V.     Two  similar  polygons  are  to  each  other 
as  the  squares  of  any  two  corresponding  sides. 


Given  the  polygons  P  and  P'  with  the  corresponding  sides 
a  and  a',  b  and  &',  c  and  c',  and  so  on. 

b2        c2 


P        az 
To  prove  that  -&  =  "^ 


so  on- 


Analysis.     Divide  the  polygons  into  similar  triangles  as 
shown  in  the  figure.     Show  that  the  ratios  of  corresponding 

a2 

triangles  are  equal,  by  showing  each  equal  to  —^.      Place  each 

a 

ratio  equal  to  this,  clear  fractions  and  add  together  the  equa- 
tions. From  this  find  the  ratio  of  the  sums  of  triangles  and 
hence  the  ratio  of  the  polygons. 


Proof. 


Ti 


r 

J*L 

s'2 
s2 


'2' 


IL   L 

T'2      t'*     s'2     a' 
Adding:     (T  +  Tl 


Why? 
Why? 

Why? 

T2)  a'2  - 


Ta'2  =  T'a2. 
TV2  =  T[a?. 
T2a'2  =  T2'a2. 


(Tf  + 


T  +  T\ 


or, 


T' 
P 


2     a' 


T'2)  a2.  Why? 
Why? 


,'2' 


State  the  theorem  proved. 


228 


PLANE  GEOMETRY 


[VII,  §  256 


256.  Problem  I.     To  construct  a  polygon  similar  to  a  given 
polygon  and  having  a  given  ratio  to  it. 


Given  the  polygon  P. 

To  construct  a  similar  polygon  n  times  as  large. 

We  shall  give  the  work  for  n  =3. 

Analysis.  Since  these  polygons  must  have  the  same  ratio 
as  the  squares  of  any  two  corresponding  sides,  we  first  find 
the  side  of  a  square  which  is  three  times  the  square  on  side 
AB,  say.  On  this  we  construct  a  polygon  similar  to  P. 
This  will  be  the  polygon  required. 

Construction. 


(a)  Construct  a  square,  AH  KB,  whose  side  equals  side 
AB  of  the  given  polygon.  Then  produce  H K,  making  HR 
=  3  HK.  Rectangle  HRBiA  will  then  equal  3  AB2. 

(6)  Construct  AiBi  a  mean  proportional  between  HR  and 
HA,  as  shown  in  the  second  figure  (see  Problem  V,  page  197), 
and  we  have  AiB2  =  3  AB2.  Therefore  A±Bi  is  the  side  of 
the  required  polygon  corresponding  to  side  AB  of  the  given 
polygon. 

On  AiBi  construct  a  polygon  similar  to  the  polygon  P. 
Construct  A  A^BiEi  similar  to  A  ABE]  then  A  BiE^Di 
similar  to  A  BED]  then  A  #iACi  similar  to  A  BDC. 


VII,  §  257] 


MENSURATION 


229 


Proof.  —  = 


p 
P' 


AB 


AB2 
ZAB 


Why? 


257.  Exercises. 

1.  A  triangular  field  has  sides  of  lengths  10  rods,  20  rods, 
and  25  rods.     Construct  a  field  half  as  large  and  of  the  same 
shape,  on  a  scale  of  5  rods  to  the  inch. 

2.  Compare  the  areas  of  two  triangles  formed  by  the 
bisector  of  an  angle  included  between  the  sides  of  a  triangle, 
whose  lengths  are  5  cm.  and  10  cm.  respectively. 

3.  Prove  that  the  diagonals  of  a  quadrilateral  divjde  the 
quadrilateral  into  four  triangles  which  form  a  proportion. 

4.  In  copying  the  map  of  a  triangular  piece  of  ground  from 
one  paper  to  another  it  was  decided  to  make  the  copy  one- 
half  as  large  as  the  map.     How  long  must  the  sides  of  the 
new  triangle  be  as  compared  with  the  first? 

5.  If  in  copying  the  map  spoken  of  in  the  above  exercise  it 
was  decided  to  make  the  new  map  one  nth  part  as  large  as  the 
first,  how  long  would  the  sides  be  as  compared  with  the  first? 

6.  Construct  the  linoleum  design  illustrated  in  the  adjoin- 
ing figure  so  that  the  square  containing  the 

star  shall  have  16  times  the  area  in  the 
illustration. 

Calculate  the  area  of  the  star  in  your 
figure. 


230 


PLANE  GEOMETRY 


[VII,  §  257 


7.  Construct  parts  of  this  field  so  that  the  area  of  the 
figures  shall  be  five  times  as  large  as  the  illustration.     Line 


out  the  figure  on  cross-section  paper  ruled  in  squares  of  the 
proper  size.  Notice  that,  to  compare  the  two  figures,  one 
of  them  should  be  turned  through  one  eighth  of  a  turn.  What 
is  the  area  of  one  of  the  octagons,  and  of  one  of  the  hexagons, 
if  the  side  of  one  of  the  squares  is  a  inches. 

8.  Construct  part  of  the  tile  border  mak- 
ing the  areas  of  the  figures  six  times  as  large 
as  shown  in  the  illustration. 

9.  Construct  part  of  the  tile  border  mak- 
ing the  area  of  your  drawing  four  times  as 
large  as  that  illustrated. 


10.  Construct  a  design  similar  to  that  of  the 
adjacent  figure  having  the  area  five  times  as  large 
as  the  area  illustrated. 


11.  Construct  parquetry  border  mak- 
ing the  areas  of  the  figures  sixteen  times 
the  areas  in  the  illustration. 

Calculate  the  areas  of  the  figures  found  in  this  design. 


VII,  §  257] 


MENSURATION 


231 


12.  Construct  this  parquetry  border  design  eight  times  as 
large  as  shown  in  the  illustration. 

Then  measure  the  lines  neces- 
sary and  calculate  the  areas  of 
the  principal  figures. 


13.  Construct  this  design  for  tile  corner  so 
that  the  square  shall  be  sixteen  times  the 
size  shown. 

Calculate  the  area. 

14.  Construct  part  of   this  parquetry 
border  so  that  the  dark  square  shall  be  one 
inch  on  the  side.    Calculate  the  areas  of  all 
the  figures  lying  within  the  large  square. 


16.  Construct  figures  similar  to  the  given 
figures  but  five  times  as  large. 


16.  Construct  the  drawing  of  shield  4  inches 
high,  3  inches  wide,  and  with  bars  f  of  an  inch  wide. 


17.  Construct  this  border  four  times  as 
wide  as  shown  in  the  illustration. 


18.  Construct  part  of  this  field 
spacing  the  parallel  lines  so  that  the 
hexagons  shall  be  just  half  an  inch 
on  a  side. 


232 


PLANE  GEOMETRY 


[VII,  §  258 


258.  Theorem  VI.  In  a  regular  polygon  the  bisectors  of 
the  angles  meet  in  a  point  equidistant  from  the  sides,  and  the 
perpendicular  bisectors  of  the  sides  meet  in  a  point  equidistant 
from  the  vertices  of  the  polygon.  These  two  points  coincide. 


Given  the  regular  polygon  A  BCD 

(a)  To  prove  that  the  bisectors  of  the  angles  meet  in  a  point 
equidistant  from  the  sides. 

(6)  To  prove  that  the  perpendicular  bisectors  of  the  sides 
meet  in  a  point  equidistant  from  the  vertices. 

Analysis,  (a)  Draw  the  bisectors  of  A  A  and  B  and  sup- 
pose them  to  meet  at  0.  Show  that  the  bisector  of  Z  C  must 
go  through  0.  (b)  Show  that  the  perpendicular  bisectors  of 
the  sides  all  go  through  0.  Let  the  student  give  the  proof. 

Definitions. 

The  lines  OA,  OB,  OC  and  so  on  (figure  above)  are  the  radii 
of  the  regular  polygon. 

The  lines  OP,  OQ,  OR  and  so  on  are  the  apothems  of  the 
regular  polygon. 

Corollary.  The  radii  of  a  regular  polygon  are  equal.  The 
apothems  of  a  regular  polygon  are  equal. 

Exercise.  A  regular  polygon  of  sixteen  sides  has  each  side 
a  inches  long  and  apothem  b  inches  long.  What  is  its  area? 
Can  you  give  any  special  values  to  a  and  6? 


vii,  §  259]  MENSURATION  233 

259.  Theorem  VII.     The  area  of  a  regular  polygon  is  equal 
to  the  product  of  its  perimeter  by  one-half  of  its  apothem. 


Given  the  regular  polygon  P,  of  area  a,  perimeter  p, 
apothem  m. 

To  prove  that  a  =  %  pm. 

Analysis.  To  show  this  we  show  that  the  polygon  can  be 
divided  into  triangles  the  sum  of  whose  areas  equals  \  pm. 
Letting  0  be  the  center  of  the  polygon,  draw  the  radii  r  to 
the  vertices,  and  prove  that  the  sum  of  the  areas  of  the  tri- 
angles formed  equals  J  pm. 

Proof.     Let  the  student  give  this  proof. 

260.  Theorem  VIII.  Two  regular  polygons  of  the  same 
number  of  sides  are  similar. 

Suggestion.  The  student  can  prove  this  by  showing  a  close 
connection  with  Art.  226. 

Corollary  1.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  are  proportional  to  their  sides,  to  their 
apothems,  to  their  radii. 

Corollary  2.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  proportional  to  the  squares  of  their  sides,  to 
the  squares  of  their  apothems,  to  the  squares  of  their  radii. 

Exercise.  The  sides  of  two  regular  polygons  are  in  the 
ratio  3  :  2,  and  the  area  of  the  first  is  54  square  inches.  What 
is  the  area  of  the  second? 


234 


PLANE  GEOMETRY 


[VII,  §  261 


261.  Areas  of  Circles. 

To  prove  theorems  on  the  circle  we  shall  connect  them 
with  theorems  on  regular  polygons,  since  those  are  the 
theorems  which  we  have  proved  and  to  which  they  are  most 
closely  related. 

This  is  a  subject  which  gave  the  ancient  Greek  mathe- 
maticians much  trouble.  Of  particular  interest  is  the  amount 
of  time  and  energy  spent  on  the  problem  of  finding  a  square 
or  other  rectilinear  figure  whose  area  exactly  equals  the  area 
of  a  given  circle.  This  was  called  the  problem  of  "squaring 
the  circle."  Closely  related  to  this  was  the  problem  of 
finding  the  ratio  of  a  circumference  to  its  diameter,  which 
attracted  an  equal  amount  of  attention. 

262.  Examine  and  discuss  the  following  figures. 


The  first  figure  is  a  circle  with  regular  polygons  inscribed 
in  it,  the  second  is  a  circle  with  regular  polygons  circumscribed 
about  it.  In  either  figure,  in  going  from  one  polygon  to  the 
next  the  number  of  sides  is  doubled. 

In  the  first  figure  the  number  of  sides  is  constantly  in- 
creasing. Is  there  any  limit  to  the  number  of  sides  that 
the  polygons  may  have? 

Notice  the  length  of  each  side.  What  is  happening  to  it 
as  you  increase  the  number  of  sides? 


Vii,  §  263]  MENSURATION  235 

Notice  the  length  of  the  perimeter.  What  is  happening 
to  it  as  you  double  the  number  of  sides  of  the  polygon? 
Could  the  perimeter  ever  coincide  with  the  circumference  of 
the  circle  by  the  process  of  doubling  the  number  of  sides? 

Notice  the  area  of  the  polygon.  What  is  happening  to 
the  area  as  you  increase  the  number  of  sides?  Can  the  area 
of  the  polygon  ever  become  the  area  of  the  circle? 

Answer  similar  questions  with  reference  to  the  second 
figure. 

So  we  have  these  facts:  In  the  inscribed  polygon  or  the 
circumscribed  polygon,  as  we  increase  the  number  of  sides, 
the  length  of  each  side  continually  decreases.  In  the  case 
of  the  inscribed  polygon  the  perimeter  and  area  are  con- 
tinually increasing.  Why?  In  the  case  of  the  circumscribed 
polygon  the  perimeter  and  area  are  continually  decreasing. 
Why? 

263.  Let  us  examine  these  changes  when  the  polygons  are 
inscribed  in  and  circumscribed  about  the  same  circle. 


In  the  first  figure  we  have  a  circle,  with  an  inscribed  and  a 
circumscribed  equilateral  triangle. 

In  the  second  figure  we  have  a  circle  with  inscribed  and 
circumscribed  regular  hexagons.  We  have  twice  as  many 
sides  as  in  the  first  figure. 

In  the  third  figure  we  have  a  circle  with  inscribed  and  cir- 
cumscribed regular  dodecagons.  We  have  again  doubled 
the  number  of  sides. 


236  PLANE  GEOMETRY  [VII,  §  264 

Suppose  that  we  continue  this  process  of  doubling  the  num- 
ber of  sides.  The  areas  of  the  inscribed  polygons  are  steadily 
increasing  and  the  areas  of  the  circumscribed  polygons  are 
steadily  decreasing.  Consequently  these  areas  are  coming 
closer  together,  or,  mathematically  speaking,  they  are  ap- 
proaching a  common  limit.  This  common  limit  is  called  the 
area  of  the  circle. 

That  the  areas  of  the  inscribed  and  circumscribed  polygons 
come  closer  together  as  we  increase  the  number  of  sides  is  also 
shown  by  the  following  table,  which  gives  the  areas.  (Here 
the  radius  of  the  circle  is  r  units.) 

Areas  of  Polygons 

Number  of  Sides.  Inscribed.  Circumscribed. 

3  1.29904  r2  5.19615  r2 

6  2.59808  r2  3.46410  r2 

12  3.00000  r2  3.21539  r2 

24  3.10583  r2  3.15966  r2 

48  3.13263  r2  3.14608  r2 

96  3.13934  r2  3.14272  r2 

192  3.14103  r2  3.14187  r2 

Exercise.  Regular  polygons  of  12  sides  are  inscribed  and 
circumscribed  about  a  circle  of  3  inches  radius.  Find  by 
the  table  the  difference  of  their  areas.  Do  the  same  for  poly- 
gons of  192  sides. 

264.  Now  study  the  perimeters  of  the  polygons.  Just  as 
in  the  case  of  areas,  the  perimeters  of  the  inscribed  polygons 
steadily  increase  and  the  perimeters  of  the  circumscribed 
polygons  steadily  decrease,  so  that  they  come  more  and  more 
nearly  together,  or  approach  a  common  limit.  This  common 
limit  is  called  the  length  of  the  circumference  of  the  circle. 

Examine  the  following  table  and  note  that  the  perimeters 
of  the  inscribed  and  circumscribed  polygons  are  becoming 
more  nearly  equal.  (Diameter  of  circle  is  d  units  in  this 
computation.) 


vii,  §  265]  MENSURATION  237 

Perimeters  oj  Polygons. 

Number  of  Sides.  Inscribed.  Circumscribed. 

3  2.59808  d  5. 19615  d 

6  3.00000  d  3.46410  d 

12  3. 10583 d  3.21539 d 

24  3.13263d  3.15966  d 

48  3.13934  d  3.14608  d 

96  3.14103d  3.14272  d 

192  3.14145  d  3.14187  d 

Exercise.  Repeat  the  preceding  exercise,  calculating 
perimeters  instead  of  areas. 

265.  The  Number  TT.     The  number  which  these  multi- 
pliers are  approaching  as  we  go  on  increasing  the  number  of 
sides  is  usually  expressed  by  the  character  TT. 

The  approximate  value  of  TT  is  3.1416,  or  less  accurately,  3}. 

This  symbol  was  first  used  in  1706.  Its  value  has  been 
found  to  more  than  700  decimal  places.  The  value  of  TT  to 
15  places  is 

TT  =  3.14159  26535  89793 

OCK 

Metius' value  is  TT  =—5.     (Error  1  :  1300000).     Divide  out 
I  lo 

the  fraction  and  see  how  many  places  in  the  value  of  TT  come 

22 

out  correctly.     Do  the  same  with  the  value  -=-. 

266.  In  the  preceding  discussion  we  have  made  the  fol- 
lowing assumptions  regarding  the  area  and  the  perimeter  of 
a  circle. 

Assumption  I.  The  area  of  a  circle  is  the  limit  approached 
by  the  area  of  a  regular  circumscribed  polygon,  or  by  the  area 
of  a  regular  inscribed  polygon,  when  the  process  of  doubling  the 
number  of  sides  is  steadily  continued. 

Assumption  II.  The  length  of  the  circumference  of  a  circle 
is  the  limit  approached  by  the  perimeter  of  the  regular  circum- 
scribed polygon,  or  the  perimeter  of  the  regular  inscribed  polygon, 


238  PLANE  GEOMETRY  [VII,  §  267 

when  the  process  of  doubling  the  number  of  sides  is  steadily 
continued. 

Corollary  1.  The  area  of  a  circle  may  be  calculated  as  closely 
as  we  please  by  calculating  the  area  of  a  regular  inscribed  or 
circumscribed  polygon  having  a  sufficiently  large  number  of 
sides. 

Corollary  2.  The  circumference  of  a  circle  may  be  calculated 
as  closely  as  we  please  by  calculating  the  perimeter  of  a  regular 
inscribed  or  circumscribed  polygon  having  a  sufficiently  large 
number  of  sides. 

267.  Definition.  A  variable  magnitude  is  said  to  approach 
a  fixed  magnitude  as  a  limit  when  the  difference  between  the 
fixed  and  variable  magnitudes  becomes  and  remains  arbi- 
trarily small  in  absolute  value. 

In  what  follows  we  shall  repeatedly  use  the  following  as- 
sumption: 

Assumption  III.  //  two  variables  are  constantly  equal  and 
each  approaches  a  limit,  the  limits  are  equal. 

Illustration.  Suppose  that  we  have  two  circles  C  and  C" 
whose  comparative  sizes  we  do  not  know. 


Suppose  that  in  circle  C  we  inscribe  a  polygon  P.  Then 
suppose  that  in  C"  we  inscribe  a  polygon  P'  which  we  know 
to  be  of  the  same  area  as  polygon  P.  Then  suppose  that 
we  double  the  number  of  sides  of  each  polygon,  and  find  that 
again  the  new  polygons  are  equal.  Suppose  that  we  con- 
tinue this  experiment  again  and  again  and  always  find 


VII,  §  268] 


MENSURATION 


239 


polygon  P'  equal  to  polygon  P.  Could  we  do  this  if  circle 
C"  were  less  than  circle  C?  You  see  readily  that  we  could  not, 
for  if  the  circle  C'  were  less  than  circle  C  it  would  stop  the 
polygon  P'  from  growing  before  the  circle  C  stopped  the 
polygon  P  from  growing.  So  that  since  we  have  assumed 
that  P'  is  always  to  remain  equal  to  P,  then  C'  could  not  be 
less  than  circle  C.  By  a  like  discussion  show  that  C'  could 
not  be  greater  than  C.  Therefore  we  assume  that  if  the 
two  variable  polygons  are  to  remain  equal,  their  limits  must 
be  equal. 

268.  Theorem  IX.     The  ratio  of  the  circumference  to  the 
diameter  of  a  circle  is  constant. 


Given  the  circles  whose  centers  are  0  and  0',  with  diam- 
eters 2  r  and  2  r',  and  circumferences  c  and  c'. 

c        c' 
To  prove  that  -^  =  ^-,- 

Analysis.  Since  we  have  studied  the  ratio  of  the  perim- 
eters of  similar  regular  polygons,  as  compared  to  the  ratio  of 
the  radii,  we  shall  inscribe  in  the  circles  two  similar  regular 
polygons  whose  perimeters  are  p  and  pf  respectively.  We 

T)  T} 

then  have  the  proportion  —  =  —,.    Double  the  number  of  sides 

of  each  of  the  polygons.     State  the  proportion  that  exists 
between  perimeters  and  radii  of  the  new  polygons.     Continue 


240  PLANE  GEOMETRY  [vn,  §  268 

this  doubling,  and  stating  ratios.     Are  these  ratios  always 
equal  to  each  other?     Will  —  continue  to  approach  —  and  — , 

continue  to  approach  —  as  we  double  the  number  of  sides?   So 

c       c' 
can  we  state  that  —  =  —  ? 

Proof.     Let  p  and  p'  be  the  perimeters  of  inscribed  equi- 
lateral triangles. 

Then         -2-  =  H-'.     why?     See  Art.  260,  Corollary  1. 

Double  the  number  of  sides  and  let  p  and  p'  be  the  lengths 
of  the  new  perimeters.     Then 

7=7-  Why? 

Continuing  to  double  the  number  of  sides, 

—  is  a  variable  which  approaches  -  as  a  limit. 

vf .  c' 

—f  is  a  variable  which  approaches  —  as  a  limit. 

-r-:7.  Why? 


State  the  theorem  proved. 

Corollary  1.     Letting  TT  express  the  ratio  -r,  we  have  c  =  TT  d. 

Corollary  2.     Since  c  =  ird,  and  d  =  2  r,  therefore  C  =  2  TT  r. 

Corollary  3.  Two  circumferences  are  proportional  to  their 
radii. 

Exercise.  Calculate  the  perimeters  of  pipes  whose  diam- 
eters, in  inches,  are  as  follows:  6,  6.5,  7,  7.5,  8.  Give  results 
in  tabular  form,  taking  the  approximate  value  of  TT,  first 

99 

==  then  3.1416. 


vii,  §  269]  MENSURATION  241 

269.  Theorem  X.     The  area  of  a  circle  is  equal  to  the  prod- 
uct of  its  circumference  by  one-half  of  its  radius. 


Given  the  circle  C  with  circumference  c  and  radius  r. 
To  prove  that  area  of  C  equals  |cr. 

Analysis.  To  prove  this  circumscribe  a  regular  polygon 
P  about  the  circle  C.  Let  ap  denote  the  area  of  the  polygon, 
and  ac  the  area  of  the  circle.  The  perimeter  of  this  polygon 
is  p  and  its  apothem  is  r,  so  that  its  area  is  \  pr. 

Therefore  ap  =  \  pr. 

Now  as  we  continually  increase  the  number  of  sides  of  the 
polygon,  p  will  approach  c,  and  aP  will  approach  ac,  as  limits. 
This  is  the  basis  of  the  proof. 

Proof.     Since  p  =  c     as  a  limit, 

then  \  pr  =  \  cr  as  a  limit. 

Also  ap  =  ac    as  a  limit. 

But  ap  =  i  pr; 

ac  =  \  cr.  Why?    Art.  267. 

Corollary  1.  Since  a  =  \  cr,  and  c  =  2irr,  the  formula  for  the 
area  of  the  circle  is 

a  =  TT  r2. 

Corollary  2.  The  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii. 

270.  Exercises. 

1.  Write  the  value  of  r  in  terms  of  a,  also  in  terms  of  c. 

2.  If  a  circle  has  a  radius  of  10  units,  find  its  circumference 
and  its  area.     If  its  radius  is  1.5;  .001;  7.12. 


242  PLANE  GEOMETRY  [vn,  §  270 

3.  If  the  area  of  a  circle  is  25  TT,  what  is  its  radius?    What  is 
its  circumference? 

Answer  the  same  questions  if  area  is  6.561.     If  .0289. 
Draw  a  circle  whose  area  shall  be  9  TT  square  inches;  10  TT 
square  inches. 

4.  Find  the  area  of  the  ring  between  the  circumference  of 
two  concentric  circles  whose  radii  are  r\  and  r2. 

5.  Using  the  answer  to  Exercise  4  as  a  formula,  find  the 
area  between  two  circumferences  whose  radii  are  4  and  10; 
whose  radii  are  .7  and  .06. 

6.  What  is  the  area  of  a  circle  which  is  tangent  to  the 
circles  mentioned  in  Exercise  4? 

What  is  this  area  if  radii  of  the  circles  are  4  and  10? 

7.  What  is  the  area  of  a  circle  which  is  inscribed  in  a  square 
of  side  s?     What  is  the  ratio  of  the  circle  to  the  square? 

8.  What  is  the  area  of  a  circle  which  is  circumscribed  about 
a  square  of  side  s? 

What  is  the  ratio  of  the  circumscribed  and  inscribed 
circles? 

9.  What  is  the  area  of  a  circle  which  can  be  inscribed  in  an 
equilateral  triangle  of  side  s? 

What  is  the  area  of  a  circle  which  can  be  circumscribed 
about  this  triangle? 

What  is  the  ratio  of  circumscribed  and 
inscribed  circles? 

10.  Compare  the  sum  of  the  areas  of  the 
small  circles  in  the  adjacent  figure  with 
the  area  of  the  whole  square  and  with  the 
area  of  its  inscribed  circle. 


11.  If  the  diameter  of  the  earth  is  taken  as  42,000,000  feet, 
what  is  its  circumference?  Use  Metius'  value  for  TT  (see 
Art.  265). 

Taking  the  radius  of  the  earth  as  3960  miles,  calculate  the 
length  of  one  degree  of  arc  on  a  meridian.  One  sixtieth  of 
this  length  is  the  nautical  mile,  or  knot, 


VII,  §  270] 


MENSURATION 


243 


12.  Explain  the  following  process  for  finding  the  diameter 
of  a  pipe  which  shall  have  just  half  the  carrying  capacity  of 
a  given  pipe. 

The  circle  represents  the  opening  of 
the  given  pipe.  A  BC  is  a  carpenter's 
square  so  placed  that  AB  =  BC. 
Then  AB  is  the  diameter  of  the  pipe 
required. 

13.  The  diameters  of  two  water 

pipes  are  3  inches  and  4  inches  respectively.  Find  the  diam- 
eter of  a  pipe  which  can  carry  as  much  water  as  both  of 
them.  Do  this  by  algebra  and  by  construction. 

14.  How  nearly  correct  is  the  following  proportion? 

Let  s  be  the  side  of  a  given  square,  and  d  the  diameter  of 
an  equal  circle. 

Then  approximately,          s  :d  =8:9. 

Calculate  d  if  s  =  10,  using  the  above 
proportion.     What  is  the  error? 

16.  Construct  the  adjoining  figure: 

Suggestion: 

Take  AB  =  BC  =  CD  =  1  inch. 

Calculate    the    area    of     the    figure 
ABEDCFA  and  of  figure  AFCDGA. 

Calculate  the  areas  using  general  numbers. 
AC  =  2  a  and  AB  =  CD  =  2  b. 

16.  Two  wheels  are  connected  by 
a  belt,  the  diameters  of  the  wheels 
being  respectively  4  feet  and  1  foot. 
If  the  large  wheel  turns  120  times  a 
minute  how  fast  does  the  small  wheel 
turn? 

17.  If  the  wheels  in  Exercise  16  have  radii  equal  to  a  feet 
and  6  feet  respectively,  and  if  the  large  wheel  turns  n  times 
a  minute,  how  fast  does  the  small  wheel  turn? 


Take 


244 


PLANE  GEOMETRY 


[VII,  §  270 


18.  Figure  ABC  is  formed  by  60°  arcs,  struck  with  radius 
AB.  Construct  the  inscribed  circle  taking  AB  equal  to  4 
inches. 

Show  that  0  is  the  center 
of  the  equilateral  triangle. 

Compute  the  radius  of  the 
circle  if  AB  =  2  a. 

Suggestion:   Show  that  line 

OA  =  —A/3,  and  that  line 


-5-  A/3, 

o 


OD  =  2a  -       A/3 
o 

=  —  -  (3  —  A/3). 

o 


Compute  the  area  of  the  circle. 
2  cm,  compute  the  radius  and  area  of  the 


Letting  a   = 
circle. 

Compute  the  lengths  of  the  arcs  AB,  BC,  CD. 

19.  Show  that  the  ring  formed  by  two  concentric  circles 
has  an  area  equal  to  the  area  of  a  circle  described  on  the  line 
which  is  the  chord  of  the  larger  circle  and  tangent  to  the 
smaller. 

20.  The  front  sprocket  wheel  has  three  times  as  many  teeth 
as  the  rear  sprocket  wheel.     The  tires  of  the  bicycle  are  28 
inches  in  diameter.     How  far  will 

the  bicycle  go  when  the  pedals 
make  one  complete  revolution? 

21.  The   wooden   rims   of   the 
wheels    in    Exercise    19    are    26 
inches  in  diameter  and  the  hubs  2 

inches  in  diameter.     The  spokes  are  fastened  tangent  to  the 
hub.    Calculate  the  length  of  a  spoke. 

Suggestion.  You  have  here  two  concentric  circles,  and 
a  line  drawn  from  a  point  in  the  outer  circle  tangent  to 
the  inner  circle.  It  is  required  to  find  the  length  of  this 
line. 


viz,  §  271]  DIVISION  OF  PERIGON  245 

PART  II— DIVISION  OF  THE  PERIGON 

271.  In  the  theorems  just  discussed  we  have  used  inscribed 
and  circumscribed  polygons.     We  shall  now  investigate  the 
construction  of  such  figures  with  rule  and  compass. 

This  necessitates  the  division  of  the  perigon  into  equal 
parts.  The  primary  division  of  the  perigon  into  equal  parts 
by  the  use  of  rule  and  compass  is  possible  only  in  certain 
special  cases.  Of  course  after  it  is  once  divided,  it  can  be 
divided  into  a  greater  number  of  parts  by  a  repeated  bisection 
of  the  angles. 

272.  Problem  II.     To  bisect  a  perigon. 

Bisect  a  perigon  on  the  same  principle  that  you  would 
bisect  any  other  angle. 

By  bisecting  the  straight  angles  thus  formed  you  can  divide 
the  perigon  into  four  equal  parts. 

By  repeatedly  bisecting  angles  you  can  divide  the  perigon 
into  2n  equal  parts. 

273.  Problem  III.     To  trisect  a  perigon. 

This  you  have  also  learned  to  do,  for  you  have  learned  to 
construct  an  equilateral  triangle.  The  angle  of  this  triangle 
is  one-sixth  of  a  perigon.  The  remaining 
angle  of  the  straight  angle  is  one-third  of  the 
perigon.  Thus : 

Angle  BOC  is  J  of  a  perigon. 

Angle  COD  is  J  of  a  perigon. 

Angle  DOB  is  J  of  a  perigon. 
The  perigon  then  can  be  divided  into  3  •  2n  equal  parts  by 
the  bisecting  of  these  angles. 

274.  Problem  IV.     To  divide  a  perigon  into  five  equal  parts. 
This  construction  is  quite  complicated,  and  we  shall  have 

to  investigate  the  construction  of  the  division  of  a  line-seg- 
ment into  what  is  called  extreme  and  mean  ratio. 


246 


PLANE  GEOMETRY 


[VII,  §  275 


275.  Definition.     A  line-segment  is  divided  in  extreme  and 
mean  ratio  when  one  of  the  parts  is  a  mean  proportional  be- 
tween the  other  part  and  the  whole  line- segment. 

In  the  figure  below,  if  P  is  to  divide  segment  AB'm  extreme 
and  mean  ratio  we  must  have 

AB  :  AP  --  AP  :  PB,    or    jf  -  p£. 

276.  Preliminary  Problem.     To  divide  a  line- segment  in 
extreme  and  mean  ratio. 


Given  the  segment  AB. 

To  divide  AB  in  extreme  and  mean  ratio,  so  that  AB  :  AP 
=  AP  :  PB. 

Construction.     Draw  the  line  BX  J_  AB  at  B.     On  BX 

lay  off  BO  equal  to  \  AB.  With  0  as  center  and  OB  as 
radius,  draw  a  circle.  Draw  line  AO  to  cut  the  circumference 
in  points  M  and  N.  On  A  B  lay  off  A  P  equal  to  AM.  Then 
P  is  the  point  sought  to  divide  the  line  internally  in  the 
required  ratio. 


Proof.     Let       s  =  the  number  of  units  in  AB. 

and  m  =  the  number  of  units  in  AP. 

s  +  m  =  the  number  of  units  in  AN. 

and  s  —  m  =  the  number  of  units  in  PB. 

Forming  a  proportion  that  we  know 
s -\-  m  _  s_ 
s         m' 


Why? 


Why?     (Art.  228,  Cor.) 


VII,  §  277]  DIVISION  OF  PERIGON  247 

Why? 


s  +  m  -  &     s-m.  ^ 


s  m 

m_  _  s  —  m 
s  m  } 
s  m 


AB      AP 


277.     We  are  now  ready  to  divide  the  perigon  into  five 
equal  parts. 


Given  the  perigon  about  0. 
To  divide  it  into  five  equal  parts. 

Construction.  Draw  the  line-segment  OA  and  divide  it 
into  extreme  and  mean  ratio,  so  that  -^-p  =  ~pT- 

With  A  as  a  center  and  OP  as  a  radius,  construct  an  arc. 

With  0  as  a  center  and  OA  as  a  radius,  construct  an  arc  in- 
tersecting the  first  arc.  Call  the  point  of  intersection  B. 
Twice  angle  AOB  is  one-fifth  of  a  perigon. 

Proof.  A  A  BO  and  A  BP  are  similar.  Why? 

A  ABP  is  isosceles.  Why? 

/  APE  =2/0.  Why? 

/  OB  A  =2/0.  Why? 

/  0  =  one-fifth  of  a  straight  angle.  Why? 

2  /  0  =  one-fifth  of  a  perigon.  Why? 
Corollary.  A  perigon  can  be  divided  into  5  •  2"  equal  parts. 


248  PLANE  GEOMETRY  [vn,  §  278 

278.  Exercises. 

0  *  ty 

1.  Solve  the  equation  —  =  -      -  (Art.  277)  for  m  in  terms 

m     s  —  m 

of  s,  and  show  that  m  =  \  (—  1  ±  \/5)  s. 

2.  If  A  B  or  s  is  4  inches  long,  find  A  P  or  m : 

(a)  By  construction. 

(b)  By  the  formula  of  Exercise  1.     Which  sign  before 
\/5  is  used? 

3.  Construct  a  regular 
pentagon. 

4.  Construct  part  of  a 
pattern,  as  in  the  figure, 
for  a   regular   dodecahe- 
dron.    This  is  one  of  the 
five  so-called  regular  sol- 
ids, bounded  by  12  faces 

which  are  regular  and  equal  pentagons. 

6.  Construct  the  adjacent  design. 

Suggestion.  Draw  a  circle  5  inches  in  diameter  and  another 
circle  4  inches  in  diameter  internally  tangent  to  the 
first  at  a  point  P.  Mark  the  vertices  of  a  regular 
pentagon  on  the  inner  circumference,  starting  at 
the  point  diametrically  opposite  to  P.  Join  each 
vertex  of  this  pentagon  with  the  second  following  vertex,  so 
forming  the  star. 

279.  External  Division  in  Extreme  and  Mean  Ratio. 

Construction.  In  the  figure,  line  AO  cuts  the  circumfer- 
ence again  at  N.  With  AN  as  radius,  and  A  as  center,  strike 
an  arc  cutting  A  B  produced  at  P'. 


_..L ^ 

p''  m  A      s 

We  can  then  show  that 


Vii,  §  280]  DIVISION  OF  PERIGON  249 


AB-.AP'  =  AP':P'B   or 

Point  Pf  is  then  said  to  divide  segment  AB  externally  in 
extreme  and  mean  ratio. 

Proof.  AM  :AB  =  AB:AN, 

m  —  s      s 

or,  —  =  —  .  Why? 

s          m 

m     m  +  s  >™    9 

Why? 


s          m 
s          m 


m     m      s 
AB      AP' 


Why? 


AP'      P'B' 

Note.  In  Exercise  1,  page  248,  two  values  of  m  are  obtained  by  solv- 
ing the  given  equation,  which  was  derived  in  the  problem  of  the  internal 
division  of  a  line-segment.  If  we  call  these  values  mi  and  w2,  we  have 

mi  =  H—  1  +  VT)  s] 

m2  =  —  i  (1  -f  \/5)  s. 

Thesejwill  be  of  opposite  signs,  because  —  1  +  V5  is  positive,  and  —  1 
—  V  5  is  negative.  If  we  lay  off  mi  to  the  right  from  A  (figure  on  p.  246) 
we  reach  point  P,  the  point  of  internal  division.  If  we  lay  off  the  nega- 
tive segment  ra2  to  the  left  from  A,  we  would  have  a  point  P',  the  point 
of  external  division. 

280.  Exercises. 

1.  Solve  —  =  — ; —  for  m  and  show  that  m  =  |  (1  ±  V  5)  s. 

m     m-{-  s 

2.  li  AB  =4  inches,  find  AP': 

(a)  By  construction. 

(b)  By  the  formula  of  Exercise  1.     Which  sign  must  be 
used? 

3.  If  the  two  values  of  m  obtained  in  Exercise  1  are  denoted 
by  mi  and  m^  respectively,  so  that 

mi  =  I  (1  -f  \/5)  s,      and      ?w2  =  |  (1  -  V5)  s, 
show  that  m2,  laid  off  to  the  right  from  A,  gives  internal 
division. 


250  PLANE  GEOMETRY  [vn,  §  281 

281.  Problem  V.     To  divide  a  perigon  into  fifteen  equal 
parts. 

1      1        2 

Analysis.     -Q-  —  T-  =  TK-  Therefore  by  subtracting  an  angle 
o      o       10 

which  is  one-fifth  of  a  perigon  from  one  that  is  one- third  of 
a  perigon  and  bisecting  the  remainder  you  will  have  an  angle 
which  is  one-fifteenth  of  a  perigon. 

Make  this  construction. 

A  perigon  can  be  divided  into  15  •  2n  equal  parts. 

282.  Problem  VI.      To  divide  a  circumference  into  equal 
parts. 

We  have  but  to  divide  the  perigon  at  the  center  into  the 
required  number  of  equal  parts,  and  the  arms  of  the  angles 
will  divide  the  circumference  and  also  the  circle  into  the  re- 
quired number  of  equal  parts. 

Therefore  the  division  of  a  circumference  into  equal  parts 
is  limited  to  the  number  of  equal  parts  into  which  we  can 
divide  the  perigon. 

283.  Problem  VII.     To  inscribe  a  regular  polygon  in  a  circle. 


Given  the  circle  whose  center  is  0. 

To  inscribe  a  regular  polygon  of  n  sides. 

Construction.  Divide  the  circumference  into  n  equal  parts, 
the  division  points  being  A,  B,  C  .  .  .  join  these  points  with 
straight  lines.  Polygon  A  BC  ...  is  the  polygon  required. 

Prove  this  by  drawing  radii  as  shown  in  the  figures,  and 
proving  congruent  triangles. 


VII,  §  284]  DIVISION  OF  PERIGON  251 

Note.  A  regular  polygon  the  number  of  whose  sides  is  a  prime 
number,  such  as  3,  5,  7,  11,  13,  17,  ....  can  be  constructed  by  rule  and 
compass  alone,  if,  and  only  if,  the  number  of  sides  can  be  expressed  by 
the  formula  22  +1,  where  n  is  an  integer.  When  n  =  0,  1,  2,  3,  the 
number  of  sides  is  3,  5,  17,  257;  these  are  the  only  polygons  with  a  prime 
number  of  sides  not  exceeding  257  which  can  be  constructed  with  rule 
and  compass  alone. 

284.  Problem  VIII.  To  circumscribe  a  regular  polygon  about 
a  circle. 


Given  the  circle  whose  center  is  0. 

To  circumscribe  a  regular  polygon  about  it. 

Construction.  Divide  the  circumference  into  n  equal  parts 
and  construct  tangents  at  the  points  of  division. 

Prove  by  drawing  in  lines  as  shown  in  the  figure,  and  prov- 
ing congruent  triangles. 

285.  Problem  IX.     To  circumscribe  a  circle  about  a  given 
regular  polygon. 

Suggestion.  Erect  the  perpendicular  bisectors  to  two  con- 
secutive sides.  The  point  where  they  meet  is  the  center  of 
the  circumscribed  circle.  For  proof  see  Art.  258. 

286.  Problem  X.    To  inscribe  a  circle  in  a  given  polygon. 

Suggestion.  Bisect  two  consecutive  angles.  The  point  of 
intersection  of  these  bisectors  is  the  center  of  the  inscribed 
circle.  For  proof  see  Art.  258. 


252  PLANE  GEOMETRY  [vn,  §  287 

287.  Theorem  XI.     In  the  same  circle,  or  in  equal  circles, 
two  sectors  have  the  same  ratio  as  their  central  angles. 


Given  two  sectors  AOB  and  BOC. 

To  prove  that  sector  AOB  is  to  sector  BOC  as  Z  AOB  is  to 
Z  BOC. 

Prove  this  by  dividing  the  angles  into  equal  parts,  and  thus 
dividing  the  sectors  into  equal  parts,  and  showing  equal  ratios. 

This  proof  holds  only  for  the  commensurable  case.  For 
the  incommensurable  case  see  Art.  292. 

Corollary.  To  find  the  area  of  a  sector  we  find  the  ratio  of  its 
angle  to  a  perigon,  and  take  that  portion  of  the  area  of  the  circle. 

In  algebraic  language,  if  a  (alpha)  is  the  number  of  radians 
in  the  angle,  then 

a_        s 
2^r  ~  772 ; 
from  which  s  =  J  r2  a. 

If  the  angle  of  the  sector  is  d  degrees,  we  shall  have  the 
proportion 

d        j_ 
360  ~7rr2' 

Then  S  =  "r2m-      . 

288.  Exercises. 

1.  It  is  interesting  to  calculate  the  perimeters  of  several 
polygons  inscribed  in  a  given  circle,  using  the  trigonometric 
functions. 


VII,  §  288] 


DIVISION  OF  PERIGON 


253 


In  the  figure,  ABC  is  an  equilateral  triangle,  OD  its 
apothem,  BD  its  half -side.  Let  the  radius  of  the  circle 
be  r.  Let  p  denote  the  perimeter  of  the  triangle. 


Then 
But 


p  = 


—  =  sin  Z  DOB 
r 


sin 


120° 


£D=rsin60°. 

p  =  6  r  sin  60°. 

Let  EF  be  one  side  of  a  regular  inscribed  polygon  of  n  sides, 
OH  its  apothem,  HF  its  half  side. 

Then  p  =  n  -2HF  =  2nHF. 

360° 


Now 


rsm 


=  rsm 


p  =  2nr  sin 


360° 

2n 
180° 

n 
180° 


n 


Calculate  p  when  n  =  4,  6,  10,  30,  60,  using  the  table  on 
page  205.  Tabulate  results.  Note  the  value  of  the  ratio  of 
p  to  r.  Compare  with  Art.  264. 

2.  Explain  the  following  construction  for  finding  where 
to  cut  off  the  corners  of  a  square 
board  to  make  it  a  regular  octa- 
gon. 

Draw  the  dotted  lines.  Bisect 
angle  ACB  by  the  line  CD.  With 
CD  as  a  radius  draw  a  circle.  This 
circle  will  mark  the  corners  of  a 
regular  octagon. 

If  the  side  of  the  square  is  10 
inches  find  the  side  of  the  octagon.  (Apply  Art.  215  to  find  AD.) 


254 


PLANE  GEOMETRY 


[VII,  §  288 


3.  If  s  is  the  side  of  a  regular  octagon  and  a  is  its  apothem, 
show  that  very  nearly, 

s  :  a  =  29  :  40. 

If  a  =  10  inches,  what  is  the  length  of  s,  using  this  pro- 
portion?    Check  by  construction. 

4.  To  draw  the  arc  of  a  circular  segment  of  given  base  A B, 
and  height  EC, 

Suggestions.    From  similar  triangles, 


=-.     (2m  =  BC.) 
h      m 


r     hm     2' h 


Give  reasons  for  these  statements. 

Assuming  values  for  a  and  h  make  the  construction,  then 
solve  for  the  value  of  r  by  algebra,  and  compare  results. 

5.  Construct  design  for 
moulding  like  that  in  the 
illustration.  Here  the 
curve  ABC  consists  of 
two  arcs  each  less  than  a 
quadrant.  To  construct 
it  draw  rectangle  ADCE} 
making  AD  =  3.5  inches 
and  CD  =  3  inches.  Point  B  is  the  center  of  this  rectangle 
and  AC  is  its  diagonal.  Locate  C\  by  drawing  the  perpen- 
dicular bisector  of  BC.  Draw  arc  BC  with  C\  as  center  and 
CiC  as  radius. 

By  changing  the  dimensions  of  the 
rectangle,  you  can  get  other  curves. 


6.  Construct  a  design  for  a  bracket. 


Follow  the  adjacent  design,  or  construct  one  of  your  own. 


VII,  §  288] 


DIVISION  OF  PERIGON 


255 


7.  Construct  the  principal  outlines  of  the 
design  in  the  adjacent  figure. 

If  the  radius  of  the  circle  which  contains 
the  "star"  is  3  inches,  calculate  the  perim- 
eter of  the  star. 


8.  Construct  a  part  of  a  rosette  of  12  leaves  so  that  OA  = 
OB  =  3  inches. 


Explain  the  construction. 

9.  Construct  a  rosette  of  six  leaves  which  shall  just  fill  a 
circle  of  3  inches  radius.  Calculate  the  length  of  AC  and 
OC  (see  figure  above)  in  this  case.  Z  AOB  is  now  60°. 

If  OA  =  r,  and  if  there  are  six  leaves,  show  that 
AC  =  J  r  V3,  and  OC  =  | 


10.  Construct  the  principal  figures  form-  Fl 

ing  this  design  for   linoleum,  making  the  h 

lines  of  your  drawing  about  6  times  as  long  Q 
as  those  of  the  illustration. 


11.  Construct  a  moulding  design  similar 
to  that  in  the  figure  and  4  inches  wide. 
Calculate  the  area  of  your  figure. 


256 


PLANE  GEOMETRY 


[VII,  §  288 


12.  Construct  a  design  for  a  moulding 
like  that  in  the  illustration.  Make  AB 
and  CD  each  half  an  inch  long,  and  use  a 
radius  of  one  inch  for  arcs  BC,  DH  and 
HF,  each  of  which  is  a  quarter  of  a 
circumference. 


13.  Construct  this  arch  so  that  your 
drawing  shall  be  4  inches  wide.  Calculate 
the  height  and  the  area  of  the  arch  when 
the  width  of  the  base  is  10  feet. 


14.  Construct 
a  rosette  of  four 
leaves  like  that  in 
the  illustration. 
If  R  is  the  radius 
of  the  large  cir- 
cle, show  that 
the  radius  of  the 
small  circles  is 


Show  that  r  =  AB. 


=  R  (\/2  -  1), 


16.  Construct  a  spiral  design 
using  semicircumferences  joined 
together  at  their  extremities. 


VII,  §  288] 


DIVISION  OF  PERIGON 


257 


16.  Construct  this  figure,  taking  AB  =  4.5  inches,  CD  = 
3.5  inches,  EF  =  1.5  inches.  Calculate  the  perimeter  and 
the  area  of  the  four-pointed  figure. 


17.  Construct  the  figure  below,  using  the  following  dimen- 
sions or  one-half  of  these  dimensions : 

OD  =  2%  inches,  OE  =  2J  inches, 

OB  =  1 J  inches,  OA  =  J  inch. 

OOi  =  5  inches,  OiF  =  5f  inches  =  0XG. 
Calculate  OFi  and  hence  HF,  and  check  by  drawing. 
Calculate  FG,  and  check  by  drawing. 


258 


PLANE  GEOMETRY 


[VII,  §  288 


18.  Construct  the  prin- 
cipal outline  of  the  figure 
below. 


19.  Construct  the  prin- 
cipal lines  of  the  figure 
below. 


20.  Construct  a  design  similar  to  the 
illustration  and  four  times  as  large. 


21.  Suggestions  for  the  construction  of  the  following  design. 
FGH  K  is  a  square  whose  side  is  6  a.  Line  CD  is  drawn  so  as 
to  cut  off  the  corner  of  the  square,  making  EF  =  FL  =  2  a. 
With  C  as  a  center  and  CD  as  a  radius  construct  arc  AD. 
Similarly  construct  the  other  arcs.  Construct  such  a  figure 
making  a  =  \  inch. 


VII,  §  288] 


DIVISION  OF  PERIGON 


259 


(1)  Show  that  the  two  arcs  meeting  at  each  corner  are 
tangent  to  each  other. 

(2)  Show  that  CD  =  2\/2a. 

(3)  Show  that  area  of  sector  ACD  =  IT  a2. 

(4)  Show  that  area  of  triangle  OCD  =  4  a2. 

(5)  Show  that  area  DRST  =  StimesOAD  =  8(4-7r)a2. 

(6)  Show  that  perimeter  DRST  =  4  TT  V2  a. 
22.  Construct  this  quadrifoil  design. 


Side  of  dotted  square  =  4  a. 
Radius  of  large  circle  =  a. 
Let  the  radius  of  small  circle  =  b. 
Prove  the  following: 

(1)  Area  of  quadrifoil  =  2  a2  (TT  +  2) . 

(2)  Perimeter  of  quadrifoil         =  4  TV  a. 

(3)  In  the  triangle  EHF,     EH  =a  +  6, 

FH  =  2  a  -  b, 
EF  =  a  -  b. 
Then  by  the  Pythagorean  theorem, 

(a  +  6)2  =  (2a-6)2+(a-6)2. 
Expanding  and  solving  for  b,  show  that 

b  =a(4-2\/3). 

(4)  Now  calculate  OH,  from  triangle  OHF; 

OH  =2a\/2(\/3-  1). 
Draw  a  figure,  using  a  =  1  inch,  and  check  your  calculation. 


260 


PLANE  GEOMETRY 


[VII,  §  288 


23.  Two  circles  intersect;  radii  are  given  n  and  r2;  distance 
between  the  centers  Oi02  =  d.     Calculate  OiA  and  AB. 


Suggestion:  Let  OiA  =  x     and     AB  =  y. 

Then  A02  =  d  -  x. 

From  triangle  O^AB]      AB2  =  n2  -  x2. 
From  triangle  A02B;      AB2  =  r22  -  (d  -  x)2. 
Therefore  n2  -  x2  =  r22  -  (d  -  x)2 

=  r22  -  d2  +  2  dx  -  x2. 
-  r,2  +  d2 


Therefore 


x  = 


2d 


Then  A B  =\/n2  -  OiA2. 

Calculate  0\A  and  A  B  when  ri  =  2  inches,  rz  =  1.5  and  d 
=  2.5  inches.  Check  by  drawing. 

24.  Draw  a  square  (four  inches  on  a  side)  and  within  it 
construct  the  figure  below. 

A,  B}  C,  D  are  the  centers  of  the  large  arcs  and  M,  N,  P,  Q 
are  the  centers  of  the  small  arcs. 


M." 


A.  a  R  a 


VII,  §  288] 


DIVISION  OF  PERIGON 


261 


If  AB  =  2  a,  show  that: 

(1)  Area  BCGHA  =TTO?. 

(2)  Area  AEFCGH  =  2  a2  (TT  -  2). 

(3)  Length  MG  =  a  (V3  -  1). 
Suggestion.     Take  R,  S  as  mid-points  of  AB,  CD. 

Then  in  triangle  BGR}          BG  =  2  a,BR  =  a. 
Therefore  RG  =  a\/3. 

Then  GS  =  2  a  -  RG  =  a  (2  -\/3). 

Then  MG  =  a  -  GS. 

(4)  Area  MGF  =  }  TT  a2  (2  -  \/3). 

(5)  Area  G##F  =  2  a2  (2  -  \/3)  (4  -  TT). 
Suggestion.     From  square  MNPQ  subtract  four  times  area 

MGF. 

25.  (1)  Construct  the  following  figure,  taking  a  =  1  inch. 


3  a 


C  a     O 


(2)  Calculate  the  perimeter  and  the  area  of  the  main 
figure  whose  corners  are  AFED. 

Answers.  Perimeter  =  (16  +  2  TT)  a. 

Area  =  (19  +  \  TT)  a2. 

(3)  Calculate  perimeter  and  area  of  figure  GH  KL. 

Answers.  Perimeter  =  ^  (12  -  \/2)  +  a  \/2, 


Area  =  —  (73 


12  \/2)  +       (25  -  2  N/2), 


262 


PLANE  GEOMETRY 


[VII,  §  288 


circles  in  a  given 


\/OOi2  -  OE2  =  x\/3. 


26.  To  inscribe  three  equal  tangent 
circle.     Trisect  the  circumference  of 
the  given  circle  at  points  A}  B,  C. 
Draw  OA,  OB,  OC.     Bisect  angle 
AOB  by  the  line  OD.     The  problem 
then  is  to  draw  a  circle  tangent  to 
the  given  circle  at  D  and  tangent  to 
OA  and  OB. 

We  can  calculate  the  position  of 
center  Oi  as  follows: 

Let  OA  =  R  and  OOi  =  2  x. 

Then  in  A  OE01}  angle  E  =  90°, 
angle  0  =  60°, 
angle  Oi  =  30°. 

Therefore  OE  =  i  00,  =  x. 

Hence  EC 

But  Oi, 

Therefore  OiJ 

Then  we  have 

OOi  +  OiD  =  R 
or  2x  +  xV3  =  R. 

Therefore  x  =  -- 

2+V3 

=  R  (2  -  V3) 

=  2R-  R  Vs. 

From  this  we  can  easily  construct  x. 
For  FG  =  2  R  and  FH  =  R  \/3,  if  we  makeGH  =  R. 
Therefore  KG  =  FG  -  FK  =  2  R  -  R  \/3  =  x. 
Now  make  OOi  =  2  KG,  thus  locating  Oi.     Using  OiD  for 
a  radius  draw  the  circle. 

27.  To  inscribe  three  equal  circles  in  a  given  circle,  the 
radius  of  the  small  circle  can  be  taken  approximately  from  the 
proportion 

r  :  R  =  13  :  28. 


VII,  §  2881 


DIVISION  OF  PERIGON 


263 


Find  the  value  of  r  in  terms  of  R  in  Exercise  26  and  find 
the  ratio  of  r  to  R.  Determine  to  how  many  decimal  places 
the  ratio  given  in  this  exercise  agrees  with  it. 

If  R  =  10  inches,  calculate  r  to  two  decimal  places  and 
check  by  construction. 

28.  Construct  the  design  here  shown,  consisting  of  a  Gothic 
arch,  two  tangent  lines,  and  inscribed  circle. 


(a)  Construct  the  arch,  making  A  B  =  3  inches. 

(b)  Draw  tangent  lines  from  point  D,  taking  D  so  that 
CH  =  HD. 

(c)  Inscribe  the  circle  in  triangle  FGD.     0  is  found  as 
the  intersection  of  HD  with  the  bisector  of  Z  FGD.    Point 
E  is  used  in  drawing  this  bisector,  GE  being  made  equal  to 
GF.     With  E  and  F  as  centers,  and  a  radius  more  than  half 
of  EFj  draw  two  intersecting  arcs,  and  through  their  in- 
tersection and  point  G  draw  line  GO. 

If  AC  =  CB  =  a,  show  that 

(1)  CD  =  2a\/3; 

(2)  DK2  =  9  a2,  or  DK  =  3  a. 

Suggestion.     Regard  DC  produced  as  a  secant  of  the  circle 
to  which  D  K  is  tangent,  and  apply  Art.  228,  Corollary. 


264 


PLANE  GEOMETRY 


[VII,  §  289 


PART  III.— INCOMMENSURABLE   CASES. 

289.  Proof  for  Incommensurable  Case  of  Theorem  I, 
Chapter  6.  //  a  pencil  of  lines  is  cut  by  a  pencil  of  parallels, 
the  corresponding  segments  are  in  proportion. 


\ 


Given  the  pencil  whose  vertex  is  V  cut  by  the  parallels 
P  and  P',  so  that  segments  a  and  b  are  incommensurable. 

To  prove  that  ~  =  ^  =  ?• 

Analysis.  Since  segments  a  and  6  are  incommensurable, 
a  unit  which  will  divide  segment  a  exactly  when  laid  off  on 
segment  b  will  leave  a  remainder.  Suppose  u  is  the  length  of 
such  a  unit,  which,  when  laid  off  on  segment  6,  will  leave 
a  remainder  MR.  A  line  drawn  through  M  parallel  to  line 
Pf  will  cut  segment  KN  commensurable  with  segment  c, 
and  LQ,  commensurable  with  e  on  the  other  two  rays  re- 
spectively, leaving  remainders  NS  and  QT.  Again,  if  we 
choose  a  unit  u'  less  than  M R,  which  is  less  than  unit  u,  we 
shall  have  segments  HMf  commensurable  with  a,  KN'  com- 
mensurable with  c,  and  LQ'  commensurable  with  e,  but  with 
remainder  in  each  case  less  than  in  the  first  case.  And  so  by 
continually  diminishing  the  unit  of  measure  we  can  make  the 
remainders  as  small  as  we  please  and  make  the  segment  HM 
approach  b}  KN  approach  d,  LQ  approach/,  as  limits 


VII,  §  290] 


INCOMMENSURABLE  CASES 


265 


Proof.     Since   segment   a   and   segment   MH   are   com- 
mensurable 

ace 

" W 


Art.  196. 


HM       KN 
As  we  decrease  the  unit  of  measure, 

HM  =  b,    KN  =  d,    LQ=f. 

a    ^_  a        c     _._  c       e    _._  e 
"  d' 


Then 


HM      b'    KN 


QL      f 


"       ^  =  «      Art.  267. 
o       rf      ; 

290.  Proof  for  Incommensurable  Case  of  Theorem  I, 
Chapter  7.  Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

D  Q 


A'-AB 


P  P'B 


Given  the  rectangles  R  and  R'  with  equal  altitudes  a,  and 
incommensurable  bases  b  and  bf  respectively. 


To  prove: 


rectangle  R       b 


rectangle  R'      b' 

Analysis.  Since  b  and  b'  are  incommensurable  bases,  a 
unit  u  which  will  divide  base  b  exactly,  when  laid  off  on  base 
6',  will  leave  a  remainder  PB,  and  will  give  segment  A  P  com- 
mensurable with  base  b.  Choose  a  new  unit  less  than  PB\ 
this  will  leave  a  remainder  P'B  less  than  PB,  making  APf 
commensurable  with  b.  By  making  the  unit  of  measure 
continually  smaller  we  can  make  the  remainder  as  small  as 
we  please,  and  thus  make  the  segment  AP  approach  base 
b'  as  its  limit.  By  drawing  perpendiculars  at  points  P,  we 
have  the  rectangles  APQD  approaching  rectangle  R'  as 


266  PLANE  GEOMETRY  [Vii,  §  291 

limit.  So  our  proof  consists  in  showing  that  by  this  process 
we  have  two  variables  constantly  equal,  and  that  the  ratios 
of  the  required  proportion  are  the  limits  of  these  equal 
variables. 

Proof.     Since  b  and  AP  are  commensurable, 

qfl      A 

n  AQ      AP' 
As  we  decrease  the  unit  of  measure, 

n  AQ  =  a  #';       base  AP  =  base  6'. 
a#   _._  a  #          6    ^  6_ 
"n~AQ~n^/;      AP  ~  b'' 

a#        L    Art  267 

n#'          6'* 

291.  Proof  for  Incommensurable  Case  of  Theorem  XI, 
Chapter  7. 

In  the  same  circle,  two  sectors  have  the  same  ratio  as  their 
central  angles. 


Given  the  sectors  AOB  and  BOC  in  the  circle  whose  center 
is  0,  A  AOB  and  BOC  being  incommensurable. 

sector  AOB      angle  AOB 
lo  prove:  $ector  BQC  =  mgle  BQC- 

Analysis.  A  unit  of  angle  ut  which  will  divide  Z  AOB 
exactly,  when  laid  off  on  Z  BOC  will  leave  a  remainder  Z 
POC.  Let  the  student  complete  this  analysis,  leading  to  the 
following  proof. 


VII,  §  292]  SUMMARY  267 

Proof.     Since  A  AOB  and  BOP  are  commensurable, 

sector  AOB  _  Z  AOB 

sector  BOP  ~  Z  BOP' 
As  we  decrease  the  unit  of  measure, 

sector  BOP  =  sector  BOC]    Z  BOP  =  Z  BOC. 

sector  AOB  ^  sector  AOB ^    Z  AOB  ^  Z  AOB 
•'*      sector  BOP  ~  sector  £OC'   Z  £OP  ~  Z  J50C* 

sector  AOB       Z 


sector  BOC 


Art.  267. 


292.  Summary  of  Chapter  7. 
Part  I — Areas  of  Plane  Figures. 

Rectilinear  figures. 

Theorem  I.  Two  rectangles  having  equal  altitudes  are  to  each  other 
as  their  bases. 

Corollary  1.  Two  rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Corollary  2.  Two  parallelograms  having  equal  altitudes  are  to  each 
other  as  their  bases,  and  having  equal  bases  are  to  each  other  as  their 
altitudes. 

Corollary  3.  Two  triangles  having  equal  altitudes  are  to  each  other 
as  their  bases  and  having  equal  bases  are  to  each  other  as  their  altitudes. 

Theorem  II.  Two  rectangles  are  to  each  other  as  the  product  of 
(the  numerical  measure  of)  their  bases  and  altitudes. 

Theorem  III.  The  area  of  a  rectangle  is  equal  to  the  product  of  its 
base  by  its  altitude. 

This  means  that  the  number  which  represents  the  square  units  in 
the  area  is  equal  to  the  product  of  the  numbers  which  represent  the 
linear  units  in  the  base  and  altitude. 

Corollary  1 .  The  area  of  a  parallelogram  equals  the  product  of  base 
by  altitude. 

Corollary  2.  The  area  of  a  triangle  equals  half  the  product  of  base 
by  altitude. 

Corollary  3.  The  area  of  a  trapezoid  equals  half  the  product  of  the 
sum  of  the  bases  by  the  altitude. 

Theorem  IV.  Two  similar  triangles  are  to  each  other  as  the  squares 
of  any  two  corresponding  sides,  or  as  the  squares  of  any  two  corre- 
sponding altitudes. 

Theorem  V.  Two  similar  polygons  are  to  each  other  as  the  squares 
of  any  two  corresponding  sides. 


268  PLANE  GEOMETRY  [VII.  §  292 

Problem  I.  To  construct  a  polygon  similar  to  a  given  polygon  and 
having  a  given  ratio  to  it. 

Definition.  Regular  Polygons.  Radii  of  Regular  Polygons. 
Apothems. 

Theorem  VI.  In  a  regular  polygon  the  bisectors  of  the  angles  meet 
in  a  point  equidistant  from  the  sides,  and  the  perpendicular  bisectors 
of  the  sides  meet  in  a  point  equidistant  from  the  vertices. 

Corollary.  The  radii  of  a  regular  polygon  are  equal.  The  apothems 
are  equal. 

Theorem  VII.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  its  perimeter  by  its  apothem. 

Theorem  VIII.  Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 

Corollary  1.  The  perimeters  of  regular  polygons  of  the  same  number 
of  sides  are  to  each  other  as  their  apothems,  as  their  radii,  to  their 
sides. 

Corollary  2.     The  areas  of  regular  polygons  are  to  each  other  as  the 
squares  of  their  apothems,  of  their  radii,  of  their  sides. 
Circles. 

Definition  of  Inscribed  and  Circumscribed  Polygons. 

Theory  of  Limits. 

Assumption  I.  The  area  of  a  circle  is  the  limit  approached  by  the 
area  of  a  regular  circumscribed  polygon,  or  by  the  area  of  a  regular 
inscribed  polygon,  when  the  process  of  doubling  the  number  of  sides 
is  steadily  continued. 

Assumption  II.  The  length  of  the  circumference  of  a  circle  is  the 
limit  approached  by  the  perimeter  of  a  regular  circumscribed  polygon, 
or  the  perimeter  of  a  regular  inscribed  polygon  as  the  process  of  doubling 
the  number  of  sides  is  steadily  continued. 

Assumption  III.  If  while  approaching  their  limits,  two  variables 
are  constantly  equal,  their  limits  are  equal. 

Theorem  IX.  The  ratio  of  the  circumference  of  a  circle  to  its  diam- 
eter is  constant. 

Corollary  1.     Letting  TT  express  the  ratio  ^,  we  have  c  =  tr  d. 

Corollary  2.  Since  c  =  tr  d  and  d  =  2  r,  c  =  2  irr. 

Corollary  3.  Two  circumferences  are  proportional  to  their  radii. 

Theorem  X.  The  area  of  a  circle  is  equal  to  half  the  product  of  cir- 
cumference by  radius. 

Corollary  1.  The  area  of  a  circle  =  IT  r2. 

Corollary  2.  The  areas  of  circles  are  to  each  other  as  the  squares  of 
their  radii. 


vii,  §  292]  SUMMARY  269 

Part  II — Division  of  a  Perigon. 

Problem  II.     To  bisect  a  perigon. 

Corollary.     To  divide  a  perigon  into  2  •  2n  equal  parts. 

Problem  III.     To  trisect  a  perigon. 

Corollary.     To  divide  a  polygon  into  3  •  2n  equal  parts. 

Problem  IV.     To  divide  a  perigon  into  five  equal  parts. 

Preliminary  Problem.     To  divide  a  line  in  extreme  and  mean  ratio. 

Corollary.     To  divide  a  perigon  into  5  •  2n  equal  parts. 
External  Division  in  Extreme  and  Mean  Ratio. 

Problem  V.     To  divide  a  perigon  into  fifteen  equal  parts. 

Corollary.     To  divide  a  perigon  into  15  •  2n  equal  parts. 

Problem  VI.     To  divide  a  circumference  into  equal  parts. 

Problem  VII.     To  inscribe  a  regular  polygon  in  a  given  circle. 

Problem  VIII.    To  circumscribe  a  regular  polygon  about  a  given 
circle. 

Problem  IX.     To  circumscribe  a  circle  about  a  given  polygon. 

Problem  X.     To  inscribe  a  circle  in  a  given  polygon. 

Theorem  XI.     In  the  same  or  equal  circles  sectors  have  the  same 
ratio  as  their  central  angles. 

Corollary.     If  s  represents  the  number  of  square  units  in  the  area  of 
a  sector  and  a  the  number  of  radians  in  its  angle,  then 

s  =  |ar2. 

If  the  angle  of  the  sector  in  degrees  is  d,  then 

,   d 
8 -^380- 

Part  III — Incommensurable  Cases. 

1.  Into  what  parts  is  this  chapter  divided? 

2.  Are  there  any  theorems  that  have  been  proved  without  the  use  of 
preceding  theorems? 

3.  State  the  theorems  that  depend  directly  upon  the  theorems  of 
Chapter  II.     State  those  that  depend  directly  on  theorems  proved  in 
each  of  the  successive  chapters. 

4.  State  the  formula  for  the  area  of  a  circle.     Then  state  the  theorem 
from  which  it  is  derived.     Then  state  the  theorems  that  are  immediately 
necessary  for  the  proof  of  this  theorem,  then  the  theorems  that  are  im- 
mediately necessary  for  the  proof  of  these  theorems  and  so  on  until 
you  come  to  theorems  that  do  not  depend  upon  other  theorems  for 
their  proof. 

It  would  be  well  to  indicate  this  chain  of  theorems  by  a  diagram 
showing  their  connection,  after  the  order  of  a  family  tree. 


270 


APPENDIX 


APPENDIX.     SOME  GEOMETRICAL  PARADOXES. 

The  need  for  accurately  drawn  figures  and  careful  proofs  will  be 
emphasized  by  a  study  of  the  following  paradoxes.  Read  each  one 
carefully;  if  you  see  nothing  wrong,  draw  an  accurate  figure  of  your 
own;  this  will  explain  the  fallacy  in  the  first  four  cases. 

(A).  From  a  point  not  on  a  line  two  perpendiculars  can  be  drawn 
to  the  line. 


Draw  two  circles  intersecting  in  P,  and  draw  the  diameters  PA  and 
PB.     Draw  AB  meeting  the  circles  in  Q  and  R  respectively. 
Then     Z  PRA  =  90°  and  Z  PQB  =  90°.   (Inscribed  in  semicircles.) 

PQ  J_  AB  and  PR  J_  AB. 
(B).    To  show  that  a  right  angle  equals  an  obtuse  angle. 


Draw  n  ABCD. 

Draw  BE  =  BC,  forming  obtuse  Z  ABE.     Draw  DE. 

Draw  J_  bisectors  of  CD  and  ED,  and  produce  them  to  meet  at  O. 

Draw  OA,  OB,  OC,  OD,  OE. 

Now  prove  A  OBC  ^  A  OBE  as  follows: 

(1)  Side  OB  is  common. 

(2)  BC  =  BE,  by  construction. 

(3)  OC  =  OE,  because  each  is  equal  to  OD.     (Art.  35.) 

A  OBC  ^  A  OBE. 
Z  OBC  =  Z  OBE. 
Subtracting  Z  OB  A  from  each  of  these;     Z  ABC  =  Z  ABE. 


PARADOXES  271 

(C).    To  prove  part  of  an  angle  equal  to  the  whole  angle. 


Draw  rt.  A  ABC,  with  A  B  and  C  less  than  60°. 

On  BC,  construct  equilateral  A  BCD. 

On  CD  lay  off  CP  =  CA. 

Let  X  =  mid-point  of  AB. 

Draw  PX  and  produce  it  to  meet  CB  at  Q.     Draw  AQ. 

Draw  J_  bisectors  of  AQ  and  PQ,  and  produce  them  to  meet  at  O. 

Draw  OC,  OA,  OP,  OQ. 

Then      OQ  =  OA  =  OP.       Also     CA  =  CP  and  CO  =  CO. 

A  AOC  &  A  POC.    .'.         Z  AGO  =  Z  PCO. 
But  Z  A  CO  is  part  of  Z  PCO. 
This  paradox  is  also  involved  in  (B). 
(D).    To  prove  that  every  triangle  is  isosceles. 


Given  A  ABC.  To  prove  AB  =  AC. 

Draw  bisector  of  Z  A  and  the  J_  bisector  of  side  BC',  produce  these 
lines  to  meet  at  D. 


272  APPENDIX 

Case  I.     Suppose  D  to  fall  within  the  triangle. 
Draw  DC  and  DB. 
Draw  DE  ±  AC  and  DF  J_  AB. 
A  ADE  &  A  ADF]  :.  AE  =  AF. 

A  EDC  ^  A  FDB,  since  #D  =  FD  and  CD  =  £Z>.     /.  EC  =  FB. 
.'.  AE  +  EC  =  AF  +  FB,  or,  AC  =  AB. 
Case  II.     Suppose  D  to  fall  without  the  triangle. 
Making  a  construction  similar  to  that  in  Case  I  and  using  congruent 
triangles,  prove  AE  =  AF  and  CE  =  BF. 
.'.     AE  -  CE  =  AF  -  BF,  or,  AB  =  AC. 
(E).    To  prove  part  of  a  line-segment  equal  to  the  whole  segment. 


Given  segment  AB,  and  on  it  a  point  X  between  A  and  B. 
To  prove  AX  =  AB. 

On  AX  as  chord  draw  a  circle,  and  from  B  draw  #<7  making  /.  ABC 
=  Z.  XCA.     (Notice  that  /.  XCA  is  an  inscribed  angle  and  will  be  the 
same  wherever  BC  meets  the  circle.) 
Draw     CP  _L  AB. 

A  ABC  is  similar  to  A  ACX.     (Equiangular.) 
A  ABC  :  A  ACX  =  BC2  :  CX*.     (Art.  265.) 
Also       A  ABC  :  A  ACX  =  AB  :  AX.     (Common  altitude.) 

BC2:AB  =  CX2:AX. 

But        BC*  =  AC2  +  A&  -  2  AB  •  AP',     (Art.  119.) 
and        CX2  =  AC2  +  AX2  -2  AX  -  AP. 

AC2+AB*-2AB>AP      AC2  +  AX*  -  2  AZ  •  AP, 
AB  AX 


AC2  -  Aff  •  AX  _  AC2  -AB-AX 
AB  AX 


PARADOXES  273 

The  numerators  of  these  fractions  are  equal. 
AB  =  AX. 

Here  the  figure  is  correctly  drawn  and  all  the  reasoning  is  correct, 
except  the  last  step. 

If  the  numerators  of  two  equal  fractions  are  equal  we  can  infer  that 
the  denominators  are  equal,  except  when  both  numerators  are  zero.     Two 

fractions  like  —  and  -r  are  equal  because  both  are  equal  to  zero,  regard- 
less of  whether  a  equalsj)  or  not.     Now  in  our  fractions  above 

AC2  -  AB  -  AX  =  0, 
because  from  similar  ^  ABC  and  AXC, 

AX  :AC  =  AC  :  AB,  or  AC2  =  AX  -  AB. 

Therefore  the  equality  of  our  two  fractions  tells  us  nothing  about 
AB  and  AX. 


INDEX 


PAGE 

Abscissa 202 

Altitude,  of  A,  O ,  or  trape- 

zoid 84 

Angle 11 

central 113 

complementary 16 

conjugate 17 

exterior 39 

inscribed 130 

positive,  negative 16 

re-entrant 13 

supplementary 17 

vertical 20 

Angles,  classified 12 

alternate-interior 42 

alternate-exterior 42 

corresponding 42 

Antecedent 169 

Apothem 232 

Arc 32 

major 113 

measurement  of 115 

minor 113 

Archimedes 4 

Area,  of  circle 237,  241 

of  .a 88 

of  CD 88 

of  A 88 

of  sector 252 

of  trapezoid 89 

Axioms 9 

Ease,  of  A,  o ,  or  trapezoid.  84 

Chord 112 

Circle..  34 


PAGE 

Circles,  concentric 139 

externally  tangent 139 

internally  tangent 139 

Circumference 34 

length  of 237,  239 

Commensurable  segments . .  169 

Composition 175 

and  division 175 

Conclusion 27 

Congruent  figures 22 

Consequent 169 

Converse  theorems 76 

Coordinates 160 

Corollary 16 

Degree 13 

Descartes 167 

Diameter 112 

Distance 202 

Division,  external 178 

internal 177 

harmonic 183 

in  a  proportion 175 

Equal  figures 84 

Euclid 4 

Extreme  and  Mean  Ratio .  .  246 

division  in 246-9 

Function 116 

Gothic  arch  (Ex.  13) 127 

Hypothesis 27 


Incommensurable  cases 264-7 

segments 169 


275 


276 


INDEX 


Limits 238 

assumptions 238 

Line 7 

straight 8 

Locus 148 

Mean  proportional 170 

construction  of 197 

Number,  irrational 169 

rational 169 

Ordinate 202 

Paradoxes 270-3 

Parallel... 44 

Parallelogram 67 

Pencil  of  b'nes 170 

of  parallels 170 

vertex  of 170 

Pi,  IT 15 

Plato 3 

Point 7 

Polygon 13 

concave 13 

convex 13 

diagonal  of 62 

regular 62 

Polygons,  named 62 

Postulates  for  angles 16 

for  circles 113 

for  parallels 44 

for  straight  lines 9 

Projection 106 

Proportion 168 

continued..  168 


PAGE 

extremes  in 170 

means  in 170 

Pythagoras 2 

Pythagorean  theorem 97 

Radian 13 

measure 15 

Radius,  of  circle 34 

of  polygon 232 

Ratio 168 

Reciprocal 172 

Rectangle 67 

Rhombus ,.  67 

Secant 112 

Sector 113 

Segment  (of  circle) 130 

major 130 

minor 130 

Similar  figures 185 

Slope  (of  aline) 165,214 

Solid,  geometric 7 

Square 67 

steel 26 

Square  Root  (construction) .  100 

Surface,  geometric 7 

..,  r 

Tangent 128 

Thales 1 

Trapezoid 70 

Triangle,  equilateral 25 

isosceles 25 

scalene 25 

Trigonometric  functions 204 

table  of..  205 


YB   17299 


BOOK 


OVERDUE. 

—=5= 


**=SSS  «C 


M306195 


L. 


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